Polytropic Process Questions in English

(A) Given the process law: $P^2 V = K$ (where $K$ is a constant).
Using the ideal gas equation $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting this into the given law: $(\frac{nRT}{V})^2 V = K \Rightarrow \frac{n^2 R^2 T^2}{V^2} V = K \Rightarrow T^2 V^{-1} = \text{constant}$.
This implies $T^2 \propto V$.
Since the gas is expanding, the volume $V$ increases.
As $V$ increases, $T^2$ must increase, which means the temperature $T$ of the gas increases.
The internal energy $U$ of an ideal gas is given by $U = n C_V T$.
Since $U \propto T$, as the temperature $T$ increases, the internal energy $U$ of the gas also increases continuously.

(B) For a polytropic process $PV^n = \text{constant}$, the molar heat capacity $C$ is given by the formula:
$C = C_V + \frac{R}{1-n} = \frac{R}{\gamma-1} + \frac{R}{1-n}$
According to the problem, the molar heat capacity $C$ is the arithmetic mean of $C_P$ and $C_V$:
$C = \frac{C_P + C_V}{2}$
Since $C_P = C_V + R$, we can write:
$C = \frac{(C_V + R) + C_V}{2} = \frac{2C_V + R}{2} = C_V + \frac{R}{2}$
Equating the two expressions for $C$:
$C_V + \frac{R}{1-n} = C_V + \frac{R}{2}$
Subtracting $C_V$ from both sides:
$\frac{R}{1-n} = \frac{R}{2}$
This implies:
$1 - n = 2$
$n = 1 - 2 = -1$
Therefore, the value of $n$ is $-1$.

(C) The given process equation is $PV^2 = C$.
Using the ideal gas law $PV = nRT$,we can write $P = \frac{nRT}{V}$.
Substituting this into the process equation: $(\frac{nRT}{V})V^2 = C$.
This simplifies to $nRTV = C$,or $TV = \text{constant}$ (since $nR$ and $C$ are constants).
Therefore,$T_1 V_1 = T_2 V_2$,which implies $\frac{T_1}{T_2} = \frac{V_2}{V_1}$.
If $V_2 > V_1$,then $\frac{V_2}{V_1} > 1$,which means $\frac{T_1}{T_2} > 1$,so $T_1 > T_2$ or $T_2 < T_1$.
Thus,option $C$ is correct.

(C) The equation of a polytropic process is $PV^x = \text{constant}$,which can be written as $\log P + x \log V = \text{constant}$,or $\log P = -x \log V + \text{constant}$.
Comparing this with the equation of a line $y = mx + c$,the slope $m = -x$.
For process $A$,the slope is $\tan(\theta_A) = \gamma$. Thus,$-x_A = \gamma$,so $x_A = -\gamma$.
The molar heat capacity for a polytropic process is $C = C_V + \frac{R}{1-x}$.
For process $A$,$C_A = C_V + \frac{R}{1 - (-\gamma)} = C_V + \frac{R}{1+\gamma}$. Since $\gamma > 1$,$C_A$ is a finite positive value.
For process $B$,the slope is $\tan(45^\circ) = 1$. Thus,$-x_B = 1$,so $x_B = -1$.
For process $B$,$C_B = C_V + \frac{R}{1 - (-1)} = C_V + \frac{R}{2}$.
Comparing the values:
$C_P = C_V + R$
$C_B = C_V + 0.5R$
$C_A = C_V + \frac{R}{1+\gamma}$ (where $1 < \gamma < 1.67$,so $0.37R < \frac{R}{1+\gamma} < 0.5R$)
Thus,$C_P > C_B > C_A > C_V$.

(A) For a polytropic process $PV^x = K$,the work done by the gas is given by the formula:
$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} K V^{-x} \, dV$
$W = \frac{K V_2^{1-x} - K V_1^{1-x}}{1-x}$
Since $P_1 V_1^x = K$ and $P_2 V_2^x = K$,we can write:
$W = \frac{P_2 V_2^x V_2^{1-x} - P_1 V_1^x V_1^{1-x}}{1-x} = \frac{P_2 V_2 - P_1 V_1}{1-x}$
Given $x = 3/2$,we substitute this into the equation:
$W = \frac{P_2 V_2 - P_1 V_1}{1 - 3/2} = \frac{P_2 V_2 - P_1 V_1}{-1/2}$
$W = -2(P_2 V_2 - P_1 V_1) = 2(P_1 V_1 - P_2 V_2)$
Thus,the correct option is $A$.

(A) For a monatomic ideal gas, the molar heat capacity $C$ for a polytropic process $PV^x = \text{constant}$ is given by $C = C_V + \frac{R}{1-x}$.
Given the process is $P/V = 1$, which implies $P = V^1$, or $PV^{-1} = \text{constant}$.
Comparing this with $PV^x = \text{constant}$, we get $x = -1$.
For a monatomic gas, the molar heat capacity at constant volume is $C_V = \frac{3R}{2}$.
Substituting the values into the formula: $C = \frac{3R}{2} + \frac{R}{1 - (-1)}$.
$C = \frac{3R}{2} + \frac{R}{2} = \frac{4R}{2} = 2R$.

(B) For a straight line passing through the origin in a $P-V$ diagram,the relationship is $P \propto V$,which means $P = kV$ or $\frac{P}{V} = \text{constant}$.
This can be written as $P V^{-1} = \text{constant}$.
Comparing this with the polytropic process equation $P V^x = \text{constant}$,we get $x = -1$.
The molar heat capacity $C$ for a polytropic process is given by $C = C_V + \frac{R}{1-x}$.
For a monoatomic gas,$C_V = \frac{3}{2} R$.
Substituting the values,$C = \frac{3}{2} R + \frac{R}{1 - (-1)} = \frac{3}{2} R + \frac{R}{2} = \frac{4 R}{2} = 2 R$.

(A) The ideal gas equation is given by $PV = nRT$.
From this, we can write $P = \frac{nRT}{V}$.
Given the condition for the process is $PV^2 = \text{constant}$.
Substituting the expression for $P$ into the given condition:
$\left(\frac{nRT}{V}\right) V^2 = \text{constant}$
$nRT V = \text{constant}$
Since $n$ and $R$ are constants, we get $TV = \text{constant}$, which implies $T \propto \frac{1}{V}$.
As the gas expands, the volume $V$ increases.
Since $T$ is inversely proportional to $V$, an increase in volume leads to a decrease in the temperature of the gas.

(C) From the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$.
Given the process law $VP^2 = \text{constant}$.
Substituting $P$ in the given law: $V \left(\frac{nRT}{V}\right)^2 = \text{constant}$.
This simplifies to $V \cdot \frac{n^2 R^2 T^2}{V^2} = \text{constant}$, which means $\frac{T^2}{V} = \text{constant}$.
Therefore, $\frac{T_1^2}{V_1} = \frac{T_2^2}{V_2}$.
Given $T_1 = T$, $V_1 = V$, and $V_2 = 2V$.
Substituting these values: $\frac{T^2}{V} = \frac{T_2^2}{2V}$.
$T_2^2 = 2T^2$.
$T_2 = \sqrt{2} T$.

(C) Given the adiabatic process equation: $Pv^{\frac{3}{2}} = \text{constant}$ ... $(i)$
From the ideal gas law: $PV = nRT$,we have $P = \frac{nRT}{V}$ ... (ii)
Substituting (ii) into $(i)$:
$\left(\frac{nRT}{V}\right) V^{\frac{3}{2}} = \text{constant}$
$T V^{\frac{1}{2}} = \text{constant}$
Therefore,$T_1 V_1^{\frac{1}{2}} = T_2 V_2^{\frac{1}{2}}$
Given initial temperature $T_1 = T$ and final volume $V_2 = \frac{V_1}{4}$:
$T V_1^{\frac{1}{2}} = T_2 \left(\frac{V_1}{4}\right)^{\frac{1}{2}}$
$T = T_2 \left(\frac{1}{4}\right)^{\frac{1}{2}}$
$T = T_2 \left(\frac{1}{2}\right)$
$T_2 = 2T$

(D) For a polytropic process $PV^n = \text{constant}$, the work done $W$ is given by $W = \frac{nR\Delta T}{1-n}$.
For a diatomic gas like Hydrogen, the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.
The change in internal energy is $\Delta U = nC_V\Delta T = n(\frac{5}{2}R)\Delta T$.
Given the process $PV^2 = \text{constant}$, we have $n = 2$.
The work done is $W = \frac{nR\Delta T}{1-2} = -nR\Delta T$.
The ratio of work done to the change in internal energy is $\frac{W}{\Delta U} = \frac{-nR\Delta T}{n(\frac{5}{2}R)\Delta T} = \frac{-1}{2.5} = -0.4$.

(D) For a polytropic process $PV^x = \text{constant}$, the work done $W$ is given by $W = \frac{nR(T_1 - T_2)}{x - 1}$.
Given $n = 1$ and $x = 3$.
The temperature change is $\Delta T = T_2 - T_1$, so $T_1 - T_2 = -\Delta T$.
Substituting these values into the formula:
$W = \frac{1 \cdot R \cdot (-\Delta T)}{3 - 1}$
$W = \frac{-R \Delta T}{2}$
The magnitude of the work done is $|W| = \left| \frac{-R \Delta T}{2} \right| = \frac{R \Delta T}{2}$.

(C) Given the condition $V \propto T^{2/3}$.
Using the ideal gas equation $PV = nRT$,we have $P = \frac{nRT}{V}$.
Since $V \propto T^{2/3}$,we can write $V = cT^{2/3}$,which implies $T \propto V^{3/2}$.
Substituting this into the ideal gas equation: $P \propto \frac{T}{V} \propto \frac{V^{3/2}}{V} = V^{1/2}$.
Thus,$P = kV^{1/2}$ for some constant $k$.
The work done $W$ is given by $W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} kV^{1/2} \, dV$.
$W = k \left[ \frac{V^{3/2}}{3/2} \right]_{V_1}^{V_2} = \frac{2}{3} [kV_2^{3/2} - kV_1^{3/2}] = \frac{2}{3} [P_2V_2 - P_1V_1]$.
Using $PV = nRT$,we get $W = \frac{2}{3} nR(T_2 - T_1) = \frac{2}{3} nR \Delta T$.
Given $n = 1 \ mol$,$\Delta T = 30 \ K$,and $R = 8.314 \ J/mol \cdot K$ (approximated as $8.3$ in calculation).
$W = \frac{2}{3} \times 1 \times 8.314 \times 30 = 20 \times 8.314 = 166.28 \ J$.
Rounding to the nearest option,$W = 166.2 \ J$.

(B) For a polytropic process $PV^n = \text{constant}$, the molar heat capacity is $C = C_V + \frac{R}{1-n}$.
Given $n = 3$ and $C_V = \frac{3R}{2}$ for a monoatomic gas.
Thus, $C = \frac{3R}{2} + \frac{R}{1-3} = \frac{3R}{2} - \frac{R}{2} = R$.
From the ideal gas law $PV = RT$ (for $1$ mole), we have $T = \frac{PV}{R}$.
Initial state: $T_1 = \frac{P_1 V_1}{R}$.
Final state: Since $P_1 V_1^3 = P_2 V_2^3$, we have $P_2 = P_1 \left( \frac{V_1}{V_2} \right)^3$.
$T_2 = \frac{P_2 V_2}{R} = \frac{P_1 V_1^3}{R V_2^2}$.
The heat absorbed is $Q = n C \Delta T = 1 \cdot R \cdot (T_2 - T_1)$.
$Q = R \left( \frac{P_1 V_1^3}{R V_2^2} - \frac{P_1 V_1}{R} \right) = P_1 V_1 \left( \frac{V_1^2}{V_2^2} - 1 \right)$.