Polytropic Process Questions in English

(B) For an ideal gas, the equation of state is $PV = nRT$. Given the process equation $PV^2 = K$ (where $K$ is a constant).
Substituting $P = \frac{nRT}{V}$ into the process equation, we get $(\frac{nRT}{V})V^2 = K$, which simplifies to $TV = \frac{K}{nR} = \text{constant}$.
Since $T \propto \frac{1}{V}$, as the gas expands, the volume $V$ increases.
Because $V$ increases, the temperature $T$ must decrease.
Therefore, the gas is cooled during the expansion.

(C) The $P-V$ diagram is a straight line passing through the origin,which implies $P \propto V$,or $P = kV$ (where $k$ is a constant).
This can be written as $PV^{-1} = \text{constant}$,which is of the form $PV^x = \text{constant}$ with $x = -1$.
The molar heat capacity $C$ for a polytropic process $PV^x = \text{constant}$ is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$.
For a diatomic gas,the adiabatic index $\gamma = 1.4 = \frac{7}{5}$.
Substituting the values: $C = \frac{R}{1.4 - 1} + \frac{R}{1 - (-1)} = \frac{R}{0.4} + \frac{R}{2} = 2.5R + 0.5R = 3R$.

(D) From the ideal gas equation,$PV = \mu RT$,so $P = \frac{\mu RT}{V}$.
Given the process law $VP^2 = K$ (where $K$ is a constant).
Substituting $P$ in the law: $V \left( \frac{\mu RT}{V} \right)^2 = K$.
$V \cdot \frac{\mu^2 R^2 T^2}{V^2} = K$.
$\frac{T^2}{V} = \frac{K}{\mu^2 R^2} = \text{constant}$.
This implies $T^2 \propto V$,or $T \propto \sqrt{V}$.
Therefore,$\frac{T_2}{T_1} = \sqrt{\frac{V_2}{V_1}}$.
Given $V_1 = V$,$V_2 = 2V$,and $T_1 = T$.
$\frac{T_2}{T} = \sqrt{\frac{2V}{V}} = \sqrt{2}$.
$T_2 = \sqrt{2} T$.

(B) The given process is described by the equation $PV^3 = \text{constant}$.
This is a polytropic process of the form $PV^n = \text{constant}$, where $n = 3$.
For an ideal gas, the molar heat capacity $C$ for a polytropic process is given by the formula $C = C_v + \frac{R}{1 - n}$.
For a monatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Substituting the values $C_v = \frac{3}{2}R$ and $n = 3$ into the formula:
$C = \frac{3}{2}R + \frac{R}{1 - 3}$
$C = \frac{3}{2}R + \frac{R}{-2}$
$C = \frac{3}{2}R - \frac{1}{2}R = R$.
Therefore, the heat capacity of the gas during this process is $R$.

(B) Given the relation $V = \frac{b}{T}$,we can write $VT = b$ (constant).
Using the ideal gas law $PV = nRT$,for $n = 1$ mole,we have $T = \frac{PV}{R}$.
Substituting this into the relation: $V \left( \frac{PV}{R} \right) = b$,which simplifies to $PV^2 = bR = \text{constant}$.
This is a polytropic process of the form $PV^x = \text{constant}$,where $x = 2$.
The molar heat capacity $C$ for a polytropic process is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$.
Substituting $x = 2$: $C = \frac{R}{\gamma - 1} + \frac{R}{1 - 2} = \frac{R}{\gamma - 1} - R = R \left( \frac{1 - (\gamma - 1)}{\gamma - 1} \right) = R \left( \frac{2 - \gamma}{\gamma - 1} \right)$.
The heat absorbed $Q$ is given by $Q = nC \Delta T$.
For $n = 1$,$Q = \left( \frac{2 - \gamma}{\gamma - 1} \right) R \Delta T$.

(B) Given the process equation: $VP^3 = k$ (where $k$ is a constant).
From the ideal gas law, $PV = \mu RT$, we can write $P = \frac{\mu RT}{V}$.
Substituting $P$ into the process equation: $V \left( \frac{\mu RT}{V} \right)^3 = k$.
This simplifies to $V \cdot \frac{(\mu R)^3 T^3}{V^3} = k$, which gives $\frac{T^3}{V^2} = \text{constant}$.
Therefore, $\frac{T_1^3}{V_1^2} = \frac{T_2^3}{V_2^2}$.
Given $T_1 = T$, $V_1 = V$, and $V_2 = 27V$, we substitute these values:
$\frac{T^3}{V^2} = \frac{T_2^3}{(27V)^2}$.
$\frac{T^3}{V^2} = \frac{T_2^3}{729V^2}$.
$T_2^3 = 729T^3$.
Taking the cube root of both sides, $T_2 = \sqrt[3]{729} T = 9T$.

(B) The work done by a gas is given by $W = \int P dV$.
Using the ideal gas law $PV = RT$ (for $1$ mole),we have $P = \frac{RT}{V}$.
Thus,$W = \int \frac{RT}{V} dV$.
Given the relation $V = k T^{2/3}$,we differentiate with respect to $T$:
$dV = k \cdot \frac{2}{3} T^{-1/3} dT$.
Substituting $k = \frac{V}{T^{2/3}}$ into the expression for $dV$:
$dV = \left( \frac{V}{T^{2/3}} \right) \cdot \frac{2}{3} T^{-1/3} dT = \frac{2}{3} \frac{V}{T} dT$.
Therefore,$\frac{dV}{V} = \frac{2}{3} \frac{dT}{T}$.
Substituting this into the work integral:
$W = \int_{T_1}^{T_2} R T \left( \frac{2}{3} \frac{dT}{T} \right) = \int_{T_1}^{T_2} \frac{2}{3} R dT$.
$W = \frac{2}{3} R (T_2 - T_1)$.
Given the change in temperature $\Delta T = T_2 - T_1 = 30^{\circ}C = 30 \ K$:
$W = \frac{2}{3} R (30) = 20 R$.

(A) Given the relation for the process is $PV^{2/3} = \text{constant}$.
Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting this into the given relation: $\left(\frac{nRT}{V}\right) V^{2/3} = \text{constant}$.
This simplifies to $T V^{-1/3} = \text{constant}$, or $T = \text{constant} \times V^{1/3}$.
Since the gas is expanding, the volume $V$ is increasing.
As $V$ increases, $V^{1/3}$ increases, which implies that the temperature $T$ must also increase.

(B) Given the equation of state: $P^2V = K$ (where $K$ is a constant).
From the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$.
Substituting $P$ into the given equation: $(\frac{nRT}{V})^2 V = K$.
This simplifies to: $\frac{n^2 R^2 T^2}{V^2} \cdot V = K$, which gives $\frac{n^2 R^2 T^2}{V} = K$.
Therefore, $T^2 \propto V$, or $T \propto \sqrt{V}$.
Let the initial state be $(T_1, V_1) = (T, V)$ and the final state be $(T_2, V_2) = (T', 2V)$.
Using the proportionality $\frac{T_1}{T_2} = \sqrt{\frac{V_1}{V_2}}$, we get $\frac{T}{T'} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}}$.
Thus, $T' = \sqrt{2} T$.

(C) For an adiabatic process,the given relation is $PV^{3/2} = \text{constant}$.
Using the ideal gas equation $PV = RT$,we can write $P = \frac{RT}{V}$.
Substituting this into the given relation: $\left(\frac{RT}{V}\right) V^{3/2} = \text{constant}$.
This simplifies to $TV^{1/2} = \text{constant}$.
Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.
Given $T_1 = T$ and $V_2 = \frac{V_1}{2}$.
Using the relation $T_1 V_1^{1/2} = T_2 V_2^{1/2}$,we get:
$T \cdot V_1^{1/2} = T_2 \cdot \left(\frac{V_1}{2}\right)^{1/2}$.
$T = T_2 \cdot \frac{1}{\sqrt{2}}$.
Therefore,$T_2 = \sqrt{2}T$.

(A) For a polytropic process $P V^x = \text{constant}$, the molar heat capacity $C$ is given by $C = C_V + \frac{R}{1-x}$.
Given the process law is $P/V = \text{constant}$, we can write it as $P V^{-1} = \text{constant}$.
Thus, the polytropic index $x = -1$.
For an ideal monoatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Substituting these values, $C = \frac{3}{2} R + \frac{R}{1-(-1)} = \frac{3}{2} R + \frac{R}{2} = 2 R$.
The heat supplied is given by $\Delta Q = n C \Delta T$.
Given $n = 1$ mole and $\Delta T = 2 T_0 - T_0 = T_0$.
Therefore, $\Delta Q = 1 \times (2 R) \times T_0 = 2 R T_0$.

(D) For a process $PV^m = \text{constant}$,the molar heat capacity $C$ is given by the formula: $C = C_V + \frac{R}{1-m}$.
For a diatomic gas,the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.
Given that the molar heat capacity $C = R$,we substitute the values into the formula:
$R = \frac{5}{2}R + \frac{R}{1-m}$.
Subtracting $\frac{5}{2}R$ from both sides:
$R - \frac{5}{2}R = \frac{R}{1-m}$.
$-\frac{3}{2}R = \frac{R}{1-m}$.
Dividing by $R$ (assuming $R \neq 0$):
$-\frac{3}{2} = \frac{1}{1-m}$.
Taking the reciprocal of both sides:
$-\frac{2}{3} = 1 - m$.
Rearranging for $m$:
$m = 1 + \frac{2}{3} = \frac{5}{3}$.
Therefore,the value of $m$ is $5/3$.

(B) The given process is $\frac{P}{V^2} = a$,which is a polytropic process of the form $PV^n = K$,where $n = -2$.
The work done in a polytropic process for $n \neq 1$ is given by the formula:
$W = \frac{nR\Delta T}{1-n}$
Here,$n = -2$,$n_{mole} = 1$,and $\Delta T = T_2 - T_1$.
Substituting these values into the formula:
$W = \frac{1 \cdot R(T_2 - T_1)}{1 - (-2)}$
$W = \frac{R(T_2 - T_1)}{1 + 2}$
$W = \frac{1}{3}R(T_2 - T_1)$

(D) The given process is $\frac{PT^2}{V} = \text{constant}$.
For a diatomic gas, the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.
Using the ideal gas equation $PV = nRT$, we have $T = \frac{PV}{nR}$, which implies $T \propto PV$.
Substituting $T$ into the given process equation:
$\frac{P(PV)^2}{V} = \text{constant} \implies \frac{P^3V^2}{V} = \text{constant} \implies P^3V = \text{constant}$.
Taking the cube root, we get $PV^{1/3} = \text{constant}$.
This is a polytropic process of the form $PV^x = \text{constant}$, where $x = 1/3$.
The molar heat capacity $C$ for a polytropic process is given by $C = C_V + \frac{R}{1-x}$.
Substituting the values: $C = \frac{5}{2}R + \frac{R}{1 - 1/3} = \frac{5}{2}R + \frac{R}{2/3} = \frac{5}{2}R + \frac{3}{2}R = \frac{8}{2}R = 4R$.

(D) From the ideal gas equation, $PV = nRT$, we have $P = \frac{nRT}{V}$.
Substituting this into the given process equation $PV^2 = K$ (where $K$ is a constant):
$(\frac{nRT}{V})V^2 = K$
$nRTV = K$
$TV = \frac{K}{nR} = \text{constant}$.
Since $T \propto \frac{1}{V}$, as the gas expands, the volume $V$ increases.
Because $V$ increases, the temperature $T$ must decrease to keep the product $TV$ constant.
Therefore, the gas undergoes cooling during this expansion.
Thus, the correct option is $D$.

(D) Given the condition $VP^2 = \text{constant}$.
From the ideal gas equation, $PV = nRT$, so $V = nRT/P$.
Substituting $V$ into the given condition: $(nRT/P) \cdot P^2 = \text{constant} \implies TP = \text{constant}$.
Thus, $P \propto 1/T$, which represents a rectangular hyperbola on a $P-T$ diagram.
Since $VP^2 = \text{constant}$, we have $P^2 \propto 1/V$, which means $P \propto 1/\sqrt{V}$.
If the volume $V$ becomes $4V$, then the new pressure $P'$ is $P' \propto 1/\sqrt{4V} = (1/2) \cdot (1/\sqrt{V}) = P/2$.
Therefore, both statements $(A)$ and $(B)$ are correct.

(B) Given: $n = 2$, $V_f = 2V_i$, $T_i = 300\, K$, process $P/V = k$ (polytropic index $x = -1$).
Using $PV = nRT$ and $P = kV$, we get $kV^2 = nRT$, so $T \propto V^2$.
$T_f = T_i (V_f/V_i)^2 = 300 \times (2)^2 = 1200\, K$.
Change in temperature $\Delta T = T_f - T_i = 1200 - 300 = 900\, K$. (Option $A$ is true).
Work done $W = \int P dV = \int_V^{2V} kV dV = k [V^2/2]_V^{2V} = k/2 (4V^2 - V^2) = 3/2 kV^2 = 3/2 PV = 3/2 nRT$.
At initial state, $W_i = 3/2 nRT_i = 3/2 \times 2 \times R \times 300 = 900\, R$. (Option $D$ is true).
Change in internal energy $\Delta U = n C_v \Delta T = 2 \times (3/2 R) \times 900 = 2700\, R$.
Heat supplied $\Delta Q = \Delta U + W = 2700\, R + 900\, R = 3600\, R$. (Option $C$ is true).
Since $\Delta Q = 3600\, R$, option $B$ is false.

(D) For a polytropic process,the molar heat capacity $C$ is given by the relation: $C = C_v + \frac{R}{1 - n}$.
Rearranging the terms to solve for $n$:
$C - C_v = \frac{R}{1 - n}$
$1 - n = \frac{R}{C - C_v}$
$n = 1 - \frac{R}{C - C_v}$
Using Mayer's relation $R = C_p - C_v$:
$n = \frac{C - C_v - (C_p - C_v)}{C - C_v}$
$n = \frac{C - C_v - C_p + C_v}{C - C_v}$
$n = \frac{C - C_p}{C - C_v}$.

(C) For a polytropic process $PV^x = \text{constant}$, the work done is given by $W = \frac{nR \Delta T}{1-x}$.
Here, $x = 2$, so $W = \frac{nR \Delta T}{1-2} = -nR \Delta T$.
For a diatomic gas like $H_2$, the change in internal energy is $\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R \right) \Delta T = 2.5 nR \Delta T$.
The ratio of work done to the change in internal energy is $\frac{W}{\Delta U} = \frac{-nR \Delta T}{2.5 nR \Delta T} = -\frac{1}{2.5} = -0.4$.

(B) Given the condition $V \propto T^{2/3}$,we can write $V = k T^{2/3}$.
From the ideal gas law $PV = nRT$,we have $P = \frac{nRT}{V}$.
Substituting $V$,we get $P = \frac{nRT}{k T^{2/3}} = \frac{nR}{k} T^{1/3}$.
Since $T \propto V^{3/2}$,we have $P \propto (V^{3/2})^{1/3} = V^{1/2}$.
This is a polytropic process $PV^x = \text{constant}$ where $P V^{-1/2} = \text{constant}$,so $x = -1/2$.
The work done in a polytropic process is given by $W = \frac{nR \Delta T}{1-x}$.
Here $n = 1 \ mol$,$\Delta T = 30 \ K$,$R = 1.99 \ cal/mol-K$,and $x = -1/2$.
$W = \frac{1 \times 1.99 \times 30}{1 - (-1/2)} = \frac{59.7}{1.5} = 39.8 \ cal$.
Converting to Joules $(1 \ cal \approx 4.184 \ J)$: $W = 39.8 \times 4.184 \approx 166.5 \ J$.
Rounding to the nearest integer,we get $167 \ J$.

(B) For a polytropic process $TV^n = \text{constant}$, the molar heat capacity is given by $C = C_V + \frac{R}{1-n} = \frac{3R}{2} + \frac{R}{1-n}$.
Given $\Delta Q = n_{m} C \Delta T$, where $n_{m} = 0.5 \text{ mole}$ and $\Delta T = 227^{\circ}\text{C} - 27^{\circ}\text{C} = 200 \text{ K}$.
Total heat $\Delta Q = 15 \times 10^3 \text{ J}$.
$15000 = 0.5 \times \left( \frac{3R}{2} + \frac{R}{1-n} \right) \times 200$.
$15000 = 100 \times \left( \frac{3R}{2} + \frac{R}{1-n} \right) \implies 150 = \frac{3R}{2} + \frac{R}{1-n}$.
Taking $R = \frac{25}{3} \text{ J/mol K}$, we get $150 = \frac{3}{2} \times \frac{25}{3} + \frac{25/3}{1-n} = 12.5 + \frac{25}{3(1-n)}$.
$137.5 = \frac{25}{3(1-n)} \implies 5.5 = \frac{1}{3(1-n)}$.
$16.5(1-n) = 1 \implies 1 - n = \frac{1}{16.5} = \frac{2}{33}$.
$n = 1 - \frac{2}{33} = \frac{31}{33}$.
Wait, re-evaluating the process equation $TV^n = \text{constant}$. The standard polytropic form is $PV^x = \text{constant}$, which implies $TV^{x-1} = \text{constant}$. Thus $n = x-1$. Using $C = \frac{R}{\gamma-1} + \frac{R}{1-x} = \frac{3R}{2} + \frac{R}{1-x}$.
From $150 = 12.5 + \frac{25}{3(1-x)}$, we get $1-x = \frac{25}{3 \times 137.5} = \frac{25}{412.5} = \frac{2}{33}$.
$x = 1 - \frac{2}{33} = \frac{31}{33}$. Since $n = x-1$, $n = \frac{31}{33} - 1 = \frac{-2}{33}$.

(B) The molar heat capacity for a polytropic process $PV^n = \text{constant}$ is given by $C = C_v + \frac{R}{1-n}$.
For a monoatomic gas, $C_v = \frac{3}{2}R$.
From the given $P-V$ graph, the process is a straight line passing through the origin, so $P \propto V$, which means $P = kV$ or $PV^{-1} = \text{constant}$.
Comparing this with $PV^n = \text{constant}$, we get $n = -1$.
Substituting the values into the formula:
$C = \frac{3}{2}R + \frac{R}{1 - (-1)}$
$C = \frac{3}{2}R + \frac{R}{2}$
$C = \frac{3R + R}{2} = \frac{4R}{2} = 2R$.
Thus, the molar heat capacity is $2R$.

(C) Given the process $VT = \text{constant}$.
Using the ideal gas law $PV = nRT$,we have $T = \frac{PV}{nR}$.
Substituting this into the process equation: $V \left( \frac{PV}{nR} \right) = \text{constant} \implies PV^2 = \text{constant}$.
This is a polytropic process $PV^x = \text{constant}$ with $x = 2$.
For a diatomic gas,the molar heat capacity at constant volume is $C_V = \frac{5R}{2}$.
The molar heat capacity for a polytropic process is $C = C_V + \frac{R}{1-x}$.
$C = \frac{5R}{2} + \frac{R}{1-2} = \frac{5R}{2} - R = \frac{3R}{2} \text{ J/mol-K}$.
The work done in a polytropic process is $W = \frac{nR(T_2 - T_1)}{1-x}$.
Given $n = 2$,$T_1 = 300 \text{ K}$,$T_2 = 500 \text{ K}$,and $x = 2$:
$W = \frac{2R(500 - 300)}{1-2} = \frac{2R(200)}{-1} = -400R \text{ J}$.
Thus,option $C$ is correct.

(C) Given the process equation: $TP^{-1/3} = \text{constant}$.
Using the ideal gas law $PV = nRT$, we have $T \propto PV$, so $PV \cdot P^{-1/3} = \text{constant}$, which simplifies to $P^{2/3}V = \text{constant}$.
Raising both sides to the power of $3/2$, we get $PV^{3/2} = \text{constant}$.
This is a polytropic process $PV^x = \text{constant}$ where $x = 3/2$.
For a monoatomic gas, the adiabatic index $\gamma = 5/3$.
The molar heat capacity $C$ for a polytropic process is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$.
Substituting the values: $C = \frac{R}{5/3 - 1} + \frac{R}{1 - 3/2} = \frac{R}{2/3} + \frac{R}{-1/2} = \frac{3R}{2} - 2R = -\frac{R}{2}$.
Since the molar heat capacity $C$ is negative, from the relation $\Delta Q = nC\Delta T$, if heat is given to the gas $(\Delta Q > 0)$, then $\Delta T$ must be negative $(\Delta T < 0)$.
Therefore, the temperature of the gas will decrease.

(D) The work done in a polytropic process $PV^n = \text{constant}$ is given by the formula:
$W = \frac{P_i V_i - P_f V_f}{n - 1}$
For a fixed initial and final state, the work done depends on the value of $n$.
If we consider the magnitude of work done for a given change in volume, the work done is zero when $n = \infty$ (isochoric process, $\Delta V = 0$).
Thus, the work done is minimum (zero) for $n = \infty$.

(B) Given the relation $V = KT^{2/3}$.
Using the ideal gas law $PV = nRT$, we have $T = \frac{PV}{nR}$.
Substituting $T$ into the given relation: $V = K \left( \frac{PV}{nR} \right)^{2/3}$.
Raising both sides to the power of $3/2$: $V^{3/2} = K^{3/2} \frac{PV}{nR}$.
Rearranging the terms: $V^{3/2} / V = \frac{K^{3/2}}{nR} P \Rightarrow V^{1/2} = \frac{K^{3/2}}{nR} P$.
This can be written as $P V^{-1/2} = \text{constant}$.
Comparing this with the polytropic process equation $PV^x = \text{constant}$, we get $x = -1/2$.
The work done in a polytropic process is given by $W = \frac{nR \Delta T}{1-x}$.
Substituting the values $n = 1$, $\Delta T = 30$, and $x = -1/2$:
$W = \frac{1 \times R \times 30}{1 - (-1/2)} = \frac{30R}{3/2} = 20R$.

(B) For a polytropic process, the equation is $P V^x = \text{constant}$.
At point $A$, $P_1 = 16 P_0$ and $V_1 = V_0$.
At point $B$, $P_2 = P_0$ and $V_2 = 2 V_0$.
Substituting these values into the polytropic equation: $P_1 V_1^x = P_2 V_2^x$.
$16 P_0 (V_0)^x = P_0 (2 V_0)^x$.
$16 = 2^x \implies 2^4 = 2^x \implies x = 4$.
The work done in a polytropic process is given by $W = \frac{P_1 V_1 - P_2 V_2}{x - 1}$.
$W = \frac{(16 P_0)(V_0) - (P_0)(2 V_0)}{4 - 1}$.
$W = \frac{16 P_0 V_0 - 2 P_0 V_0}{3} = \frac{14}{3} P_0 V_0$.

(D) Given the process equation: $PV^2 = \text{constant}$.
Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting this into the process equation: $\left(\frac{nRT}{V}\right) V^2 = \text{constant}$.
This simplifies to $nRT \cdot V = \text{constant}$.
Since $n$ and $R$ are constants, we get $TV = \text{constant}$, or $T \propto \frac{1}{V}$.
The graph of $T$ versus $V$ is a rectangular hyperbola, not a parabola or a straight line.
For an expansion, $V$ increases. Since $T \propto \frac{1}{V}$, as $V$ increases, $T$ must decrease.
Therefore, such an expansion is possible only with cooling.

(A) Given the process relation: $VT = K$.
Using the ideal gas law $PV = nRT$,we have $T = \frac{PV}{nR}$.
Substituting $T$ into the process equation: $V \left( \frac{PV}{nR} \right) = K$.
This simplifies to $PV^2 = nRK$,which is a polytropic process of the form $PV^x = \text{constant}$,where $x = 2$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_v = \frac{3}{2} R$.
The molar heat capacity $C$ for a polytropic process is given by $C = C_v + \frac{R}{1-x}$.
Substituting the values: $C = \frac{3}{2} R + \frac{R}{1-2} = \frac{3}{2} R - R = \frac{1}{2} R$.
The heat absorbed by $n$ moles of gas is $\Delta Q = n C \Delta T$.
Given $n = 1$,we get $\Delta Q = 1 \times \left( \frac{1}{2} R \right) \times \Delta T = \frac{1}{2} R \Delta T$.

(B) The work done by an ideal gas is given by $W = \int_{V_1}^{V_2} P \, dV$.
Given the process law $P/V = K$ (where $K$ is a constant), we have $P = KV$.
Substituting this into the work integral: $W = \int_{V_1}^{V_2} KV \, dV = \frac{1}{2} K(V_2^2 - V_1^2)$.
From the ideal gas equation $PV = RT$, and substituting $P = KV$, we get $(KV)V = RT$, which implies $KV^2 = RT$.
Therefore, $K V_2^2 = RT_2$ and $K V_1^2 = RT_1$.
Substituting these into the work expression: $W = \frac{1}{2} (KV_2^2 - KV_1^2) = \frac{1}{2} (RT_2 - RT_1) = \frac{R}{2}(T_2 - T_1)$.

(B) For a monoatomic gas,the molar heat capacity $C$ for a process $P = KV$ is calculated as follows:
Given $P = KV$,we know $PV = nRT$. Substituting $P = KV$,we get $KV^2 = nRT$,so $V^2 \propto T$.
For a polytropic process $PV^x = \text{constant}$,we have $T V^{x-1} = \text{constant}$.
Since $V^2 \propto T$,we have $T V^{-2} = \text{constant}$,which implies $x - 1 = -2$,so $x = -1$.
The molar heat capacity $C$ is given by $C = C_v + \frac{R}{1-x} = \frac{3}{2}R + \frac{R}{1-(-1)} = \frac{3}{2}R + \frac{R}{2} = 2R$.
The heat supplied is $Q = nC\Delta T = 1 \times (2R) \times (2T_0 - T_0) = 2RT_0$.

(B) The work done by a gas is given by $W = \int P \, dV$.
Using the ideal gas law $PV = RT$ (for $1 \, \text{mole}$),we have $P = \frac{RT}{V}$.
Given the relation $V = KT^{2/3} \dots (i)$.
Differentiating with respect to $T$,we get $dV = K \cdot \frac{2}{3} T^{-1/3} \, dT \dots (ii)$.
Dividing equation $(ii)$ by $(i)$,we get $\frac{dV}{V} = \frac{2}{3} \frac{dT}{T}$.
Substituting $P = \frac{RT}{V}$ into the work integral: $W = \int \frac{RT}{V} dV$.
Substituting $\frac{dV}{V} = \frac{2}{3} \frac{dT}{T}$,we get $W = \int R T \left( \frac{2}{3} \frac{dT}{T} \right) = \int \frac{2}{3} R \, dT$.
Integrating over the temperature change $\Delta T = 30^oC$,we get $W = \frac{2}{3} R \Delta T$.
$W = \frac{2}{3} \times 30R = 20R$.

(A) The given equation is $PV^n = C$,where $n = 1.4$.
This represents a polytropic process.
Since $P = \frac{C}{V^{1.4}}$,as volume $V$ increases,pressure $P$ decreases non-linearly.
This relationship $P \propto V^{-1.4}$ represents a curve that is steeper than an isothermal process $(n=1)$ and is characteristic of an adiabatic process $(n=\gamma=1.4)$.
Graph $A$ represents this inverse non-linear relationship (a hyperbola-like curve),which is the correct representation for $PV^{1.4} = C$.

(A) Given the process equation: $P^2V = \text{constant}$.
From the ideal gas law,$PV = nRT$,we have $P = \frac{nRT}{V}$.
Substituting this into the given equation:
$(\frac{nRT}{V})^2 V = \text{constant}$
$\frac{T^2}{V^2} \cdot V = \text{constant}$
$\frac{T^2}{V} = \text{constant}$.
This implies that $\frac{T_1^2}{V_1} = \frac{T_2^2}{V_2}$.
Given $T_1 = T_0$,$V_1 = V_0$,and $V_2 = 2V_0$:
$\frac{T_0^2}{V_0} = \frac{T_2^2}{2V_0}$
$T_2^2 = 2T_0^2$
$T_2 = \sqrt{2} T_0$.

(A) The given equation for the thermodynamic process is $PV^{2.5} = 0.40$.
To find the slope of the $P-V$ curve,we need to calculate the derivative $\frac{dP}{dV}$.
First,express $P$ as a function of $V$:
$P = 0.40 V^{-2.5}$
Differentiating both sides with respect to $V$:
$\frac{dP}{dV} = 0.40 \times (-2.5) \times V^{-2.5-1}$
$\frac{dP}{dV} = -1.0 \times V^{-3.5}$
Now,evaluate the slope at $V = 1 \, m^3$:
$\left(\frac{dP}{dV}\right)_{V=1} = -1.0 \times (1)^{-3.5}$
$\left(\frac{dP}{dV}\right)_{V=1} = -1 \, Pa/m^3$
Thus,the slope of the $P-V$ curve at $V = 1 \, m^3$ is $-1$.

(A) For an ideal gas,the equation of state is $PV = nRT$. Given $P = kV$ and $n = 1$,we have $kV^2 = RT$,which implies $V^2 \propto T$.
Since $P = kV$,the process is a polytropic process $PV^{-1} = k$. Thus,the polytropic index $x = -1$.
The change in internal energy is given by $\Delta U = nC_v \Delta T = 1 \cdot \frac{3R}{2} \cdot (2T_0 - T_0) = \frac{3RT_0}{2}$.
The work done in a polytropic process is $W = \frac{nR \Delta T}{1-x} = \frac{1 \cdot R \cdot (2T_0 - T_0)}{1 - (-1)} = \frac{RT_0}{2}$.
According to the first law of thermodynamics,the heat supplied is $Q = \Delta U + W = \frac{3RT_0}{2} + \frac{RT_0}{2} = 2RT_0$.

(A) Given the relation for the gas is $P^2V = \text{constant}$.
Using the ideal gas law, $PV = nRT$, we can express pressure as $P = \frac{nRT}{V}$.
Substituting this into the given relation: $\left(\frac{nRT}{V}\right)^2 V = \text{constant}$.
This simplifies to $\frac{n^2R^2T^2}{V^2} \cdot V = \text{constant}$, which implies $\frac{T^2}{V} = \text{constant}$.
For two states, we have $\frac{T_1^2}{V_1} = \frac{T_2^2}{V_2}$.
Given $T_1 = T_0$, $V_1 = V_0$, and $V_2 = 2V_0$, we substitute these values:
$\frac{T_0^2}{V_0} = \frac{T_2^2}{2V_0}$.
Canceling $V_0$ from both sides, we get $T_2^2 = 2T_0^2$.
Taking the square root, the final temperature is $T_2 = \sqrt{2} T_0$.

(D) Given the law for the gas is $P^2V = C$ (constant).
From the ideal gas equation,we have $PV = nRT$,which implies $P = \frac{nRT}{V}$.
Substituting $P$ into the given law: $(\frac{nRT}{V})^2 V = C$.
This simplifies to $\frac{n^2 R^2 T^2}{V^2} \cdot V = C$,or $\frac{T^2}{V} = \text{constant}$.
Therefore,$T^2 \propto V$.
For the initial state,we have $T_0^2 \propto V_0$.
For the final state,let the temperature be $T_f$ and volume be $3V_0$,so $T_f^2 \propto 3V_0$.
Dividing the two relations: $\frac{T_f^2}{T_0^2} = \frac{3V_0}{V_0} = 3$.
Taking the square root on both sides,we get $T_f = \sqrt{3} T_0$.

(A) The given process is $P = aV^2$, which can be written as $PV^{-2} = a$.
Comparing this with the polytropic process equation $PV^x = \text{constant}$, we get $x = -2$.
The molar heat capacity $C$ for a polytropic process is given by $C = C_V + \frac{R}{1-x}$.
For a diatomic gas, the molar heat capacity at constant volume is $C_V = \frac{5R}{2}$.
Substituting the values, we get $C = \frac{5R}{2} + \frac{R}{1 - (-2)}$.
$C = \frac{5R}{2} + \frac{R}{3} = \frac{15R + 2R}{6} = \frac{17R}{6}$.

(N/A) The work done in a polytropic process $PV^n = K$ is given by $W = \frac{P_1V_1 - P_2V_2}{n-1}$.
Here, $n = 1/2$. So, $W = \frac{P_1V_1 - P_2V_2}{1/2 - 1} = \frac{P_1V_1 - P_2V_2}{-1/2} = 2(P_2V_2 - P_1V_1)$.
Since $P_1V_1^{1/2} = P_2V_2^{1/2}$, we have $P_2 = P_1(V_1/V_2)^{1/2}$.
Thus, $W = 2[P_1(V_1/V_2)^{1/2}V_2 - P_1V_1] = 2P_1V_1[(V_2/V_1)^{1/2} - 1]$.
$(b)$ Using the ideal gas law $PV = RT$ (for $1$ mole), $P = RT/V$.
Substituting into $PV^{1/2} = K$, we get $(RT/V)V^{1/2} = K$, so $TV^{-1/2} = \text{constant}$.
Therefore, $T_1V_1^{-1/2} = T_2V_2^{-1/2} \implies \frac{T_1}{T_2} = (\frac{V_1}{V_2})^{1/2}$.
Given $V_2 = 2V_1$, $\frac{T_1}{T_2} = (\frac{V_1}{2V_1})^{1/2} = \frac{1}{\sqrt{2}}$.
$(c)$ From the first law of thermodynamics, $Q = \Delta U + W$.
$\Delta U = \frac{3}{2}R(T_2 - T_1)$.
Since $T_2 = \sqrt{2}T_1$, $\Delta U = \frac{3}{2}R(\sqrt{2}T_1 - T_1) = \frac{3}{2}RT_1(\sqrt{2} - 1)$.
$W = 2(P_2V_2 - P_1V_1) = 2R(T_2 - T_1) = 2RT_1(\sqrt{2} - 1)$.
$Q = \frac{3}{2}RT_1(\sqrt{2} - 1) + 2RT_1(\sqrt{2} - 1) = \frac{7}{2}RT_1(\sqrt{2} - 1)$.

(N/A) From the graph,the process follows $PV^{\frac{1}{2}} = K$ (constant).
$(a)$ The work done $W$ for the process from state $1$ to $2$ is given by:
$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{K}{\sqrt{V}} \, dV = K \left[ \frac{V^{\frac{1}{2}}}{\frac{1}{2}} \right]_{V_1}^{V_2} = 2K(\sqrt{V_2} - \sqrt{V_1})$.
Since $K = P_1 V_1^{\frac{1}{2}}$,we have $W = 2P_1 V_1^{\frac{1}{2}}(\sqrt{V_2} - \sqrt{V_1}) = 2P_1 V_1 (\sqrt{\frac{V_2}{V_1}} - 1)$.
Given $V_2 = 2V_1$,$W = 2P_1 V_1 (\sqrt{2} - 1)$.
$(b)$ From the ideal gas equation $PV = nRT$,and $P = \frac{K}{\sqrt{V}}$,we get $T = \frac{PV}{nR} = \frac{K\sqrt{V}}{nR}$.
Thus,$\frac{T_1}{T_2} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V_1}{2V_1}} = \frac{1}{\sqrt{2}}$.
$(c)$ Heat supplied $Q = \Delta U + W$.
$\Delta U = nC_v \Delta T = 1 \cdot \frac{3}{2}R(T_2 - T_1) = \frac{3}{2}(P_2 V_2 - P_1 V_1)$.
Since $P_2 V_2^{\frac{1}{2}} = P_1 V_1^{\frac{1}{2}}$,$P_2 = P_1 \sqrt{\frac{V_1}{V_2}} = \frac{P_1}{\sqrt{2}}$.
$P_2 V_2 = \frac{P_1}{\sqrt{2}} (2V_1) = P_1 V_1 \sqrt{2}$.
$\Delta U = \frac{3}{2} P_1 V_1 (\sqrt{2} - 1)$.
$Q = \frac{3}{2} P_1 V_1 (\sqrt{2} - 1) + 2 P_1 V_1 (\sqrt{2} - 1) = \frac{7}{2} P_1 V_1 (\sqrt{2} - 1)$.

(D) For a polytropic process, the equation is $PV^{x} = \text{constant}$.
Given the process equation is $PV^{1/2} = \text{constant}$.
Using the ideal gas equation $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting this into the process equation:
$\left(\frac{nRT}{V}\right) V^{1/2} = \text{constant}$
$T V^{-1} V^{1/2} = \text{constant}$
$T V^{-1/2} = \text{constant}$
Thus, $T_{1} V_{1}^{-1/2} = T_{2} V_{2}^{-1/2}$.
Rearranging for the ratio of temperatures:
$\frac{T_{2}}{T_{1}} = \left(\frac{V_{2}}{V_{1}}\right)^{-1/2} = \left(\frac{V_{1}}{V_{2}}\right)^{1/2}$.
Given $V_{2} = 2V_{1}$, we have $\frac{V_{1}}{V_{2}} = \frac{1}{2}$.
Therefore, $\frac{T_{2}}{T_{1}} = \left(\frac{1}{2}\right)^{1/2} = \frac{1}{\sqrt{2}}$.
Hence, the correct option is $D$.

(A) The process equation is given by $p V^2 = \beta$.
Since the gas is ideal,it obeys the ideal gas equation $p V = n R T$.
From the ideal gas equation,we can write $p = \frac{n R T}{V}$.
Substituting this into the process equation: $\left(\frac{n R T}{V}\right) V^2 = \beta$,which simplifies to $n R T V = \beta$.
Given values:
$n = 0.02 \, mol$
$R = 8.31 \, J \cdot K^{-1} \cdot mol^{-1}$
$T = 20^{\circ} C = 293 \, K$
$V = 1500 \, cm^3 = 1500 \times 10^{-6} \, m^3 = 1.5 \times 10^{-3} \, m^3$.
Now,calculate $\beta$:
$\beta = n R T V = (0.02) \times (8.31) \times (293) \times (1.5 \times 10^{-3})$.
$\beta = 0.02 \times 8.31 \times 293 \times 0.0015 \approx 0.0729 \, Pa \cdot m^6$.
Rounding to the nearest given option,$\beta \approx 7.5 \times 10^{-2} \, Pa \cdot m^6$.

(B) Given the process equation: $p V^2 = C$.
For an ideal gas,we know that $p V = n R T$,which implies $p = \frac{n R T}{V}$.
Substituting this into the process equation:
$\left(\frac{n R T}{V}\right) V^2 = C$
$n R T V = C$
Since $n, R,$ and $C$ are constants,we have $T V = \text{constant}$,or $T \propto \frac{1}{V}$.
If $V_2 > V_1$,then according to the relation $T \propto \frac{1}{V}$,the temperature $T_2$ must be less than $T_1$ $(T_2 < T_1)$.
Therefore,option $(b)$ is correct.

(D) For a polytropic process $p V^x = C$,the molar heat capacity $C_{process}$ is given by the formula $C_{process} = C_V + \frac{R}{1 - x}$.
Here,the given process is $p V^3 = C$,so $x = 3$.
For a monoatomic ideal gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Substituting these values into the formula:
$C_{process} = \frac{3}{2} R + \frac{R}{1 - 3}$
$C_{process} = \frac{3}{2} R + \frac{R}{-2}$
$C_{process} = \frac{3}{2} R - \frac{1}{2} R = R$.
Thus,the heat capacity of the gas during the process is $R$.

(C) The process equation is $p V^{\alpha} = K$ (constant).
From the ideal gas law,$pV = nRT$,so $p = \frac{nRT}{V}$.
Substituting this into the process equation: $\left(\frac{nRT}{V}\right) V^{\alpha} = K \Rightarrow T V^{\alpha-1} = \text{constant}$.
Differentiating with respect to $T$,we relate the change in temperature $\Delta T$ to the change in volume $\Delta V$.
The work done in a polytropic process $p V^{\alpha} = K$ is given by $\Delta W = \int p dV = \frac{p_f V_f - p_i V_i}{1-\alpha} = \frac{nR \Delta T}{1-\alpha}$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
We know $\Delta Q = C \Delta T$ and $\Delta U = C_V \Delta T$.
Substituting these into the first law equation: $C \Delta T = C_V \Delta T + \frac{nR \Delta T}{1-\alpha}$.
Dividing by $\Delta T$,we get the molar heat capacity $C = C_V + \frac{nR}{1-\alpha}$.

(C) In a polytropic process,the molar specific heat $C$ is given by the relation $C = C_V + \frac{R}{1-N}$,where $C_V$ is the molar specific heat at constant volume and $N$ is the polytropic index.
We know that for an ideal gas,$C_V = \frac{R}{\gamma-1}$.
Substituting this into the expression,we get $C = \frac{R}{\gamma-1} + \frac{R}{1-N}$.
This can be rewritten as $C = \frac{R}{\gamma-1} - \frac{R}{N-1}$.
Therefore,the correct option is $C$.

(B) The given process is a polytropic process defined by $P \propto V^{-2}$, which can be written as $P V^2 = \text{constant}$.
Comparing this with the general polytropic equation $P V^n = \text{constant}$, we get the polytropic index $n = 2$.
The work done by an ideal gas in a polytropic process is given by the formula $W = \frac{\mu R (T_1 - T_2)}{n - 1}$, or equivalently $W = \frac{\mu R \Delta T}{1 - n}$.
Given: number of moles $\mu = 2$, initial temperature $T_1 = 300 \, K$, final temperature $T_2 = 400 \, K$, and $n = 2$.
Substituting these values into the formula:
$W = \frac{2 \times R \times (400 - 300)}{1 - 2}$
$W = \frac{2 \times R \times 100}{-1}$
$W = -200 \, R$.
Thus, the work done by the gas is $-200 \, R$.

(B) For a polytropic process,the relation between pressure $P$ and volume $V$ is given by $PV^n = \text{constant}$.
Given that the process is parabolic,$P \propto V^2$,which implies $P = aV^2$. However,looking at the standard form $PV^n = K$,we rewrite this as $P V^{-2} = K$,so $n = -2$.
For an ideal diatomic gas,the adiabatic index $\gamma = 1.4 = \frac{7}{5}$.
The molar specific heat $C$ for a polytropic process is given by $C = \frac{R}{\gamma - 1} + \frac{R}{1 - n}$.
Substituting the values: $C = \frac{R}{\frac{7}{5} - 1} + \frac{R}{1 - (-2)} = \frac{R}{\frac{2}{5}} + \frac{R}{3} = \frac{5R}{2} + \frac{R}{3}$.
Calculating the sum: $C = \frac{15R + 2R}{6} = \frac{17R}{6}$.

(D) For a polytropic process $PV^n = K$, the molar heat capacity is given by $C = C_v + \frac{R}{1-n}$.
For a diatomic gas, $C_v = \frac{5}{2}R$. Given $n = 1.3$, $C = \frac{5}{2}R + \frac{R}{1-1.3} = 2.5R - 3.33R = -0.83R$.
Since $C < 0$, the gas absorbs heat when it expands (as $dQ = nCdT$ and $dT > 0$ for expansion in this process, but here $C$ is negative, implying heat is released or absorbed depending on the sign of $dT$).
Actually, for $1 < n < \gamma$ (where $\gamma = 1.4$ for diatomic gas), the gas absorbs heat during expansion and its temperature decreases.
Statement $A$: The gas absorbs heat during expansion. This is correct.
Statement $B$: The gas cools down during expansion. This is correct as $T \propto V^{1-n} = V^{-0.3}$.
Statement $C$: Work done by the gas is positive during expansion, so work done by the surroundings is negative. This is correct.
Since all statements $A$, $B$, and $C$ are correct, the incorrect statement is none of these.