Mix Examples-Thermodynamics Questions in English

(D) Given,the ratio of specific heat capacities $\gamma = 1.5 = 3/2$.
For an adiabatic process,$P_i V_i^\gamma = P_f V_f^\gamma$. Given $V_f = 2V_i$ and $P_f = P_1$,we have $P_{i,ad} V_i^{1.5} = P_1 (2V_i)^{1.5}$.
Thus,$P_{i,ad} = P_1 (2)^{1.5} = P_1 (2\sqrt{2})$.
For an isothermal process,$P_i V_i = P_f V_f$. Given $V_f = 2V_i$ and $P_f = P_2$,we have $P_{i,iso} V_i = P_2 (2V_i)$.
Thus,$P_{i,iso} = 2P_2$.
Given $P_1 = P_2$,the ratio of initial pressures is $\frac{P_{i,ad}}{P_{i,iso}} = \frac{P_1 (2\sqrt{2})}{2P_1} = \frac{2\sqrt{2}}{2} = \sqrt{2} : 1$.

(C) Let the initial state be $A$ with $(P_0, V_0, T_0)$.
In the isothermal process $A \to B$, the temperature remains constant, so the state at $B$ is $(P', 8V_0, T_0)$.
In the adiabatic process $B \to C$, the gas is compressed to volume $V_0$. The adiabatic relation is $T V^{\gamma-1} = \text{constant}$.
For the process $B \to C$:
$T_0 (8V_0)^{\gamma-1} = T'' (V_0)^{\gamma-1}$
Given $\gamma = 4/3$, so $\gamma - 1 = 1/3$.
$T_0 (8V_0)^{1/3} = T'' (V_0)^{1/3}$
$T_0 \cdot 8^{1/3} = T''$
$T'' = 2 T_0$
The average kinetic energy per molecule is given by $U_{av} = \frac{3}{2} k_B T$, which implies $U_{av} \propto T$.
Therefore, the ratio of the average kinetic energy in the final state to the initial state is:
$\frac{(U_{av})_f}{(U_{av})_i} = \frac{T''}{T_0} = \frac{2 T_0}{T_0} = 2$.

(A) For an adiabatic process,the relation between temperature and volume is $T V^{\gamma-1} = \text{constant}$.
Step $1$: Adiabatic expansion from $V_0$ to $2 V_0$.
$T_0 V_0^{\gamma-1} = T_1 (2 V_0)^{\gamma-1} \Rightarrow T_1 = T_0 \left(\frac{1}{2}\right)^{\gamma-1}$.
Step $2$: Isothermal expansion from $2 V_0$ to $5 V_0$.
Since the process is isothermal,the temperature remains constant at $T_1$.
Step $3$: Adiabatic compression from $5 V_0$ to $V_f$ such that the final temperature is $T_0$.
$T_1 (5 V_0)^{\gamma-1} = T_0 (V_f)^{\gamma-1}$.
Substituting $T_1$ from Step $1$:
$T_0 \left(\frac{1}{2}\right)^{\gamma-1} (5 V_0)^{\gamma-1} = T_0 (V_f)^{\gamma-1}$.
$\left(\frac{5 V_0}{2}\right)^{\gamma-1} = (V_f)^{\gamma-1}$.
$V_f = 2.5 V_0$.
Comparing with $V_f = \alpha V_0$,we get $\alpha = 2.5$.

(D) Given: $n = 2$ moles. The volume is doubled from $V$ to $2V$.
For the isobaric process,the work done is $W_1 = P \Delta V = P(2V - V) = PV$.
For the isothermal process,the work done is $W_2 = nRT \ln(\frac{V_2}{V_1})$.
Since $PV = nRT$,we can substitute $nRT$ with $PV$.
Thus,$W_2 = PV \ln(\frac{2V}{V}) = PV \ln(2)$.
Comparing the two expressions,since $W_1 = PV$,we get $W_2 = W_1 \ln(2)$.

(B) For an ideal gas, the equation of state is $pV = \mu RT$, which implies $p = (\frac{\mu R}{V})T$.
In a $p-T$ diagram, lines passing through the origin represent isochoric processes (constant volume) because $p \propto T$ implies $V = \text{constant}$.
Processes $AB$ and $CD$ are straight lines passing through the origin, so they are isochoric processes.
Work done in an isochoric process is zero. Thus, $W_{AB} = 0$ and $W_{CD} = 0$.
Processes $BC$ and $DA$ are isobaric processes (constant pressure).
Work done in an isobaric process is $W = p\Delta V = \mu R \Delta T$.
For process $BC$ (at pressure $p_2$): $W_{BC} = \mu R(T_C - T_B) = 3 \times R \times (2400 - 800) = 3R \times 1600 = 4800R$.
For process $DA$ (at pressure $p_1$): $W_{DA} = \mu R(T_A - T_D) = 3 \times R \times (400 - 1200) = 3R \times (-800) = -2400R$.
The total work done in the cycle is $W = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 4800R + 0 - 2400R = 2400R$.
Substituting $R = 8.314 \, J/mol \cdot K$: $W = 2400 \times 8.314 = 19953.6 \, J \approx 20 \, kJ$.

(D) $1$. In the $p-V$ diagram,the isothermal expansion from $A$ to $B$ follows the curve $AB$. The work done by the gas is the area under the curve $AB$,which is the area $ABDE$.
$2$. The adiabatic compression from $B$ to $C$ follows the curve $BC$. The work done on the gas is the area under the curve $BC$,which is the area $BCDE$.
$3$. From the graph,it is clear that the final pressure $p_2$ at point $C$ is greater than the initial pressure $p_1$ at point $A$,so $p_2 > p_1$.
$4$. The net work done by the gas $W$ is the sum of the work done during expansion $(W_{exp} > 0)$ and the work done during compression $(W_{comp} < 0)$.
$5$. Since the area under the adiabatic curve $BC$ (which represents the magnitude of work done on the gas) is greater than the area under the isothermal curve $AB$ (which represents the magnitude of work done by the gas),the net work done $W = W_{AB} - W_{BC}$ is negative $(W < 0)$.
$6$. Therefore,$p_2 > p_1$ and $W < 0$.

(C) The process $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ is a clockwise cycle.
$1$. During $A \rightarrow B$: Pressure $p$ is constant (isobaric process).
$2$. During $B \rightarrow C$: The process follows $p \propto \frac{1}{V}$, which implies $pV = \text{constant}$, so temperature $T$ is constant (isothermal process).
$3$. During $C \rightarrow D$: Volume $V$ is constant (isochoric process).
$4$. During $D \rightarrow A$: The process follows $p \propto \frac{1}{V}$, which implies $pV = \text{constant}$, so temperature $T$ is constant (isothermal process).
Comparing these characteristics with the given options, we look for a $V-T$ diagram where:
- $A \rightarrow B$ is a process where $V \propto T$ (isobaric, since $pV = nRT \Rightarrow V = (\frac{nR}{p})T$).
- $B \rightarrow C$ is a vertical line (isothermal, $T$ is constant).
- $C \rightarrow D$ is a horizontal line (isochoric, $V$ is constant).
- $D \rightarrow A$ is a vertical line (isothermal, $T$ is constant).
Option $C$ correctly represents this cycle in the $V-T$ plane.

(A) In the $P-V$ diagram:
$1$. Process $AB$ is an isobaric process,where pressure $P$ remains constant. In a $P-T$ diagram,this is represented by a horizontal line.
$2$. Process $BC$ is an isothermal process,where temperature $T$ remains constant. In a $P-T$ diagram,this is represented by a vertical line.
$3$. Process $CD$ is an isochoric process,where volume $V$ remains constant. Since $PV = nRT$,for constant $V$,$P \propto T$. Thus,in a $P-T$ diagram,this is a straight line passing through the origin.
$4$. Process $DA$ is an adiabatic process,which is represented by a curve in the $P-T$ diagram.
Comparing these characteristics,the correct $P-T$ diagram is shown in option $(a)$.

(B) In an isothermal process,the temperature of the gas remains constant,so the gas obeys Boyle's law: $P \propto \frac{1}{V}$.
Let the initial pressure be $P$ and initial volume be $V_1$. After isothermal compression,pressure $P_2 = 2P$ and volume $V_2 = V_1/2$.
Thus,$\frac{V_1}{V_2} = 2$.
Now,the gas expands adiabatically to its original volume $V_1$. Let the final pressure be $P_3 = 0.75P$.
For an adiabatic process,$P_2 V_2^\gamma = P_3 V_1^\gamma$.
Substituting the values: $(2P) \left(\frac{V_1}{2}\right)^\gamma = (0.75P) V_1^\gamma$.
$2 \cdot \left(\frac{1}{2}\right)^\gamma = 0.75$.
$2^{1-\gamma} = \frac{3}{4} = 3 \cdot 2^{-2}$.
$2^{3-\gamma} = 3$.
Taking log on both sides: $(3-\gamma) \log 2 = \log 3$.
$3-\gamma = \frac{\log 3}{\log 2} \approx 1.585$.
$\gamma = 3 - 1.585 = 1.415 \approx 1.41$.

(A) According to the ideal gas law,$PV = nRT$. Since the volume $V$ of the room is constant,we have $T = \frac{PV}{nR}$.
The internal energy $U$ of an ideal gas is given by $U = nC_vT$.
Substituting the expression for $T$,we get $U = nC_v \left( \frac{PV}{nR} \right) = \left( \frac{C_v V}{R} \right) P$.
Since $C_v$,$V$,and $R$ are constants,the internal energy $U$ is directly proportional to the pressure $P$ $(U \propto P)$.
Given that the pressure $P$ increases linearly with time,the internal energy $U$ must also increase linearly with time.

(C) The heat absorbed per cycle is $Q_H = 130 \text{ cal}$.
The heat rejected per cycle is $Q_C = 30 \text{ cal}$.
The net heat converted to work per cycle is $W_{\text{net}} = (Q_H - Q_C) \times 4.2 \text{ J/cal} = (130 - 30) \times 4.2 = 100 \times 4.2 = 420 \text{ J}$.
The energy consumed to overcome friction per cycle is $W_f = 2 \text{ J}$.
The useful work delivered to the load per cycle is $W_{\text{load}} = W_{\text{net}} - W_f = 420 \text{ J} - 2 \text{ J} = 418 \text{ J}$.
The engine operates at $90 \text{ cycles per minute}$, which is $90/60 = 1.5 \text{ cycles per second}$.
The power delivered to the load is $P = W_{\text{load}} \times \text{frequency} = 418 \text{ J} \times 1.5 \text{ s}^{-1} = 627 \text{ W}$.

(C) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v} = \frac{3}{2}R$. The change in internal energy is given by $\Delta U = n C_{v} \Delta T = n \left(\frac{3}{2}R\right) \Delta T$.
Using the ideal gas law $pV = nRT$,we have $\Delta T = \frac{p_f V_f - p_i V_i}{nR}$.
Substituting the values: $\Delta U = \frac{3}{2} (p_f V_f - p_i V_i) = \frac{3}{2} (4 p_{0} V_{0} - p_{0} V_{0}) = \frac{3}{2} (3 p_{0} V_{0}) = \frac{9}{2} p_{0} V_{0}$.
The work done $W$ is the area under the $p-V$ graph,which is a trapezoid: $W = \frac{1}{2} (p_i + p_f) (V_f - V_i) = \frac{1}{2} (p_{0} + 2 p_{0}) (2 V_{0} - V_{0}) = \frac{1}{2} (3 p_{0}) (V_{0}) = \frac{3}{2} p_{0} V_{0}$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
$\Delta Q = \frac{9}{2} p_{0} V_{0} + \frac{3}{2} p_{0} V_{0} = \frac{12}{2} p_{0} V_{0} = 6 p_{0} V_{0}$.

(A) Given the equation of state: $PV = AT - BT^2$.
Since the pressure $P$ is constant,we differentiate the equation with respect to $T$:
$P dV = A dT - B(2T) dT$.
Work done $W = \int P dV$.
Substituting $P dV$ into the integral:
$W = \int_{T_1}^{T_2} (A - 2BT) dT$.
$W = A \int_{T_1}^{T_2} dT - 2B \int_{T_1}^{T_2} T dT$.
$W = A(T_2 - T_1) - 2B \left[ \frac{T^2}{2} \right]_{T_1}^{T_2}$.
$W = A(T_2 - T_1) - B(T_2^2 - T_1^2)$.

(A) For a cyclic process,the change in internal energy $\Delta U$ is zero. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$,the net heat absorbed $\Delta Q$ is equal to the net work done $\Delta W$ by the gas.
The net work done in a cyclic process is equal to the area enclosed by the cycle in the $p-V$ diagram.
The area of the triangle in the $p-V$ diagram is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
Here,the base is $(V_{1}-V_{2})$ and the height is $(p_{1}-p_{2})$.
Therefore,the net heat absorbed $\Delta Q = \frac{1}{2}(p_{1}-p_{2})(V_{1}-V_{2})$.

(D) In the given $P-T$ diagram:
$AB$: The pressure $P$ is constant,and temperature $T$ decreases. Since $PV = nRT$,if $P$ is constant and $T$ decreases,volume $V$ must decrease. Thus,$AB$ is an isobaric compression.
$BC$: The process is a vertical line,meaning temperature $T$ is constant. Since $P$ decreases,$V$ must increase to keep $PV$ constant. Thus,$BC$ is an isothermal expansion.
$CA$: The process is a line passing through the origin in a $P-T$ diagram,meaning $P \propto T$. Since $PV = nRT$,$P/T = nR/V$,so $V$ must be constant. Thus,$CA$ is an isochoric process.
Comparing these characteristics with the given options,the correct $P-V$ diagram is represented by option $D$.

(A, B, C, D) In a $p-v$ diagram:
$1$. The process $1 \rightarrow 2$ is a horizontal line,meaning pressure $p$ is constant. Thus,it is an isobaric process.
$2$. The process $3 \rightarrow 1$ is a vertical line,meaning volume $v$ is constant. Thus,it is an isochoric process.
$3$. The process $2 \rightarrow 3$ is a curve,which represents an adiabatic process in this cycle.
$4$. For any complete cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = \Delta U + W$. Since $\Delta U = 0$,$Q = W$. The work done $W$ in a cyclic process is equal to the area enclosed by the cycle,which is non-zero. Therefore,both work done and heat absorbed are non-zero.
Thus,statements $A$,$B$,$C$,and $D$ are all correct.

(B) For a diatomic gas,the total heat given is $\Delta Q_{total} = \Delta Q_{AB} + \Delta Q_{BC}$.
$1$. For process $AB$ (Isochoric process):
$\Delta Q_{AB} = n C_V \Delta T = n \left( \frac{R}{\gamma - 1} \right) \Delta T = \frac{P_f V_f - P_i V_i}{\gamma - 1}$
Given $n = 1$ mole,diatomic gas $\gamma = 1.4 = 7/5$.
$\Delta Q_{AB} = \frac{2 P_0 V_0 - P_0 V_0}{7/5 - 1} = \frac{P_0 V_0}{2/5} = 2.5 P_0 V_0$.
$2$. For process $BC$ (Isothermal process):
$\Delta Q_{BC} = W_{BC} = n R T \ln \left( \frac{V_f}{V_i} \right) = P_B V_B \ln \left( \frac{V_C}{V_B} \right)$
Since $P_B V_B = 2 P_0 V_0$ and $V_C = 2 V_0, V_B = V_0$:
$\Delta Q_{BC} = 2 P_0 V_0 \ln \left( \frac{2 V_0}{V_0} \right) = 2 P_0 V_0 \ln 2 = 2 P_0 V_0 (0.7) = 1.4 P_0 V_0$.
Total heat $\Delta Q_{total} = 2.5 P_0 V_0 + 1.4 P_0 V_0 = 3.9 P_0 V_0$.

(B) In a $p-V$ diagram, the slope of an adiabatic curve at any point is steeper than that of an isothermal curve at that same point.
Let the initial state be $A(p_{i}, V_{i})$.
During the reversible isothermal expansion to volume $V_{0}$, the gas moves along the isothermal curve to state $B(p_{2}, V_{0})$, where $p_{2} < p_{i}$.
During the subsequent reversible adiabatic compression from volume $V_{0}$ back to $V_{i}$, the gas moves along an adiabatic curve from state $B(p_{2}, V_{0})$ to state $C(p_{f}, V_{i})$.
Since the adiabatic curve is steeper than the isothermal curve, the path from $B$ to $C$ lies above the path from $A$ to $B$ in the $p-V$ plane.
Therefore, the final pressure $p_{f}$ at volume $V_{i}$ must be greater than the initial pressure $p_{i}$.
Thus, $p_{f} > p_{i}$.

(B, C) The internal energy $(U)$ of an ideal gas is a function of temperature $(T)$ only, given by $U = f(T)$.
In an isothermal process, the temperature of the gas remains constant $(\Delta T = 0)$.
Since the internal energy depends only on temperature, it remains constant in an isothermal process.
In an isobaric process, if the gas expands, the temperature increases (from $PV = nRT$, if $P$ is constant and $V$ increases, $T$ must increase), leading to an increase in internal energy.
Therefore, both statement $(B)$ and statement $(C)$ are true.

(A, C) For a diatomic gas,the degree of freedom $f = 5$. The internal energy is $U = \frac{n f R T}{2} = \frac{5 R}{2} (\alpha t + \frac{1}{2} \beta t^2)$.
The rate of increase in internal energy is $\frac{dU}{dt} = \frac{5 R}{2} (\alpha + \beta t)$.
Since the pressure is constant,the heat supplied is $dQ = n C_p dT$. For a diatomic gas,$C_p = \frac{7}{2} R$.
Thus,the power supplied by the heating element is $i^2 r = \frac{dQ}{dt} = n C_p \frac{dT}{dt} = 1 \times \frac{7}{2} R \times (\alpha + \beta t)$.
Therefore,the current $i = \sqrt{\frac{7 R}{2 r} (\alpha + \beta t)}$.
From the ideal gas law $PV = nRT$,since $P$ is constant,$V = \frac{nRT}{P} = \frac{R}{P} (\alpha t + \frac{1}{2} \beta t^2)$.
The position of the piston $x = \frac{V}{A} = \frac{R}{PA} (\alpha t + \frac{1}{2} \beta t^2)$.
The velocity $v = \frac{dx}{dt} = \frac{R}{PA} (\alpha + \beta t)$.
The acceleration $a = \frac{dv}{dt} = \frac{R \beta}{PA}$,which is constant. Thus,options $(a)$ and $(c)$ are correct.

(A) Let the initial state of the gas be $(P_i, V, T)$.
Step $1$: Isothermal compression.
The gas is compressed until the pressure doubles. Since the process is isothermal,$P_i V = P_f V_f$. Given $P_f = 2P_i$,we have $P_i V = (2P_i) V_f$,which gives $V_f = V/2$.
The state after isothermal compression is $(2P_i, V/2, T)$.
Step $2$: Adiabatic expansion.
The gas expands adiabatically to regain its original volume $V$. Let the final pressure be $P_f'$.
Using the adiabatic process equation $P_1 V_1^\gamma = P_2 V_2^\gamma$:
$(2P_i) (V/2)^\gamma = P_f' V^\gamma$
$P_f' = (2P_i) \left(\frac{V/2}{V}\right)^\gamma = (2P_i) \left(\frac{1}{2}\right)^\gamma = 2P_i \times 2^{-\gamma}$
$P_f' = P_i \times 2^{1-\gamma}$
Given $\gamma = 1.4$,we have $P_f' = P_i \times 2^{1-1.4} = P_i \times 2^{-0.4}$.
Wait,the problem states $2^{-1.4} = 0.38$. Let's re-evaluate: $P_f' = 2P_i \times (1/2)^{1.4} = 2P_i \times 2^{-1.4}$.
Substituting the given value: $P_f' = 2P_i \times 0.38 = 0.76 P_i$.
Therefore,the ratio of the final to initial pressure is $P_f'/P_i = 0.76: 1$.

(B) For an isothermal process,the work done is given by $W_{\text{isothermal}} = nRT \ln \left(\frac{V_2}{V_1}\right)$.
Given $n = 1$,$T = 27^{\circ} C = 300 \ K$,and $V_2/V_1 = 2$,we have $W = 1 \cdot R \cdot 300 \cdot \ln(2) \approx 300R(0.693)$.
For an adiabatic process,the work done is $W_{\text{adiabatic}} = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
For a diatomic gas,$\gamma = 1.4$.
Given $W_{\text{isothermal}} = W_{\text{adiabatic}}$,we equate the two expressions:
$\frac{1 \cdot R(300 - T_{\text{final}})}{1.4 - 1} = 300R(0.693)$.
$\frac{300 - T_{\text{final}}}{0.4} = 300(0.693)$.
$300 - T_{\text{final}} = 0.4 \cdot 300 \cdot 0.693 = 83.16$.
$T_{\text{final}} = 300 - 83.16 = 216.84 \ K$.
Converting to Celsius: $T(^{\circ} C) = 216.84 - 273.15 \approx -56.31^{\circ} C$.
Thus,the final temperature is close to $-56^{\circ} C$.

(C) The change in internal energy $\Delta U$ for an ideal gas is given by the formula: $\Delta U = n C_v \Delta T$.
Given that the process occurs at constant volume,$C_v = C_p - R$.
Substituting the given values: $C_v = 7 - 2 = 5 \text{ cal/mol}^{\circ} C$.
Number of moles $n = 10 \text{ mol}$.
Change in temperature $\Delta T = 40^{\circ} C - 30^{\circ} C = 10^{\circ} C$.
Therefore,$\Delta U = 10 \times 5 \times 10 = 500 \text{ cal}$.

(B) For a diatomic gas with rotational modes only,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). Thus,$C_v = \frac{f}{2}R = \frac{5}{2}R$.
Change in internal energy $\Delta U = nC_v \Delta T = 1 \times \frac{5}{2} \times 8.3 \times 1.2 = 24.9 \text{ J}$.
Work done by the gas $W = P \Delta V = P(A \Delta x)$.
Given $P = 10^5 \text{ Pa}$,$A = 4 \times 10^{-4} \text{ m}^2$,and $\Delta x = 25 \times 10^{-3} \text{ m}$.
$W = 10^5 \times 4 \times 10^{-4} \times 25 \times 10^{-3} = 1 \text{ J}$.
Total heat supplied $Q = \Delta U + W = 24.9 + 1 = 25.9 \text{ J}$.
Comparing with the given options,$25 \text{ J}$ is the closest value.

(D) For a diatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{7}{2}R$ and at constant volume is $C_v = \frac{5}{2}R$.
Heat supplied at constant pressure is given by $\Delta Q = nC_p \Delta T = n(\frac{7}{2}R)\Delta T$.
Change in internal energy is given by $\Delta U = nC_v \Delta T = n(\frac{5}{2}R)\Delta T$.
According to the first law of thermodynamics,work done is $\Delta W = \Delta Q - \Delta U = n(\frac{7}{2}R - \frac{5}{2}R)\Delta T = n(\frac{2}{2}R)\Delta T = nR \Delta T$.
The ratio $\Delta Q : \Delta U : \Delta W$ is $\frac{7}{2} : \frac{5}{2} : \frac{2}{2} = 7 : 5 : 2$.

$A$ is correct: The Zeroth law of thermodynamics provides the concept of temperature.
$B$ is correct: The First law of thermodynamics provides the concept of internal energy.
$C$ is incorrect: For an isothermal expansion of an ideal gas, the change in internal energy $\Delta U = 0$. According to the first law $\Delta Q = \Delta U + \Delta W$, therefore $\Delta Q = \Delta W$.
$D$ is correct: The product of an intensive variable and an extensive variable is an extensive variable (e.g., $P \times V = \text{Energy}$, which is extensive).
$E$ is incorrect: The ratio of an extensive variable to mass is an intensive property (e.g., $\text{Volume} / \text{Mass} = \text{Density}$, which is intensive).