Question

A certain amount (say, $n$ moles) of a monatomic ideal gas ($C_v = 3/2R$) of a volume $V_1$ and pressure $P_1$ is expanded against a constant external pressure $P_2$ until the pressure of the gas becomes $P_2$. The correct statement is:

• A. If the expansion is carried out adiabatically, The work done is given by $w = \frac{3}{5}V_1(P_2 - P_1)$.
• B. If the expansion is carried out adiabatically, the entropy change is given by $\Delta S = 0$.
• C. If the expansion is carried out isothermally, the work done is given by $w = -nRT_1 \ln \left( \frac{P_1}{P_2} \right)$.
• D. If the expansion is carried out isothermally, the entropy change is given by $\Delta S = -nR \ln \left( \frac{P_1}{P_2} \right)$.

Answer 1

Answer: (A)

If the expansion is carried out adiabatically, The work done is given by $w = \frac{3}{5}V_1(P_2 - P_1)$.

Answer 2

The problem describes the expansion of a monatomic ideal gas against a constant external pressure $P_2$ until the gas pressure also becomes $P_2$. This is an irreversible process. We need to evaluate each statement.

Initial State: $(P_1, V_1, T_1)$ Final State: $(P_2, V_2, T_2)$ External Pressure: $P_{ext} = P_2$ (constant) Gas Type: Monatomic ideal gas, so $C_v = \frac{3}{2}R$.

Analysis of Option (A): If the expansion is carried out adiabatically, the work done is given by $w = \frac{3}{5}V_1(P_2 - P_1)$.

1. Process: Irreversible adiabatic expansion.
2. First Law of Thermodynamics: $\Delta U = q + w$.
3. Adiabatic condition: $q = 0$. So, $\Delta U = w$.
4. Change in Internal Energy for ideal gas: $\Delta U = nC_v\Delta T = n\left(\frac{3}{2}R\right)(T_2 - T_1)$.
5. Work done against constant external pressure: $w = -P_{ext}\Delta V = -P_2(V_2 - V_1)$.
6. Equating $\Delta U$ and $w$: $n\left(\frac{3}{2}R\right)(T_2 - T_1) = -P_2(V_2 - V_1)$
7. Using Ideal Gas Law: $P_1V_1 = nRT_1 \implies T_1 = \frac{P_1V_1}{nR}$ and $P_2V_2 = nRT_2 \implies T_2 = \frac{P_2V_2}{nR}$.
8. Substitute $T_1$ and $T_2$ into the equation from step 6: $n\left(\frac{3}{2}R\right)\left(\frac{P_2V_2}{nR} - \frac{P_1V_1}{nR}\right) = -P_2V_2 + P_2V_1$ $\frac{3}{2}(P_2V_2 - P_1V_1) = -P_2V_2 + P_2V_1$ Multiply by 2: $3(P_2V_2 - P_1V_1) = -2P_2V_2 + 2P_2V_1$ $3P_2V_2 - 3P_1V_1 = -2P_2V_2 + 2P_2V_1$ Rearrange to find $V_2$: $3P_2V_2 + 2P_2V_2 = 2P_2V_1 + 3P_1V_1$ $5P_2V_2 = (2P_2 + 3P_1)V_1$ $V_2 = \frac{(2P_2 + 3P_1)V_1}{5P_2} = \frac{2}{5}V_1 + \frac{3}{5}\frac{P_1V_1}{P_2}$
9. Substitute $V_2$ back into the work equation: $w = -P_2(V_2 - V_1) = -P_2\left(\left(\frac{2}{5}V_1 + \frac{3}{5}\frac{P_1V_1}{P_2}\right) - V_1\right)$ $w = -P_2\left(\frac{2}{5}V_1 - V_1 + \frac{3}{5}\frac{P_1V_1}{P_2}\right)$ $w = -P_2\left(-\frac{3}{5}V_1 + \frac{3}{5}\frac{P_1V_1}{P_2}\right)$ $w = \frac{3}{5}P_2V_1 - \frac{3}{5}P_1V_1$ $w = \frac{3}{5}V_1(P_2 - P_1)$ This statement is correct. For an expansion, $P_1 > P_2$, so $(P_2 - P_1)$ is negative, making $w$ negative, consistent with work done by the system.

Analysis of Option (B): If the expansion is carried out adiabatically, the entropy change is given by $\Delta S = 0$.

• The process is an expansion against a constant external pressure, which is an irreversible process.
• For an irreversible adiabatic process, the entropy of the system increases, i.e., $\Delta S_{sys} > 0$.
• $\Delta S = 0$ only for a reversible adiabatic process. This statement is incorrect.

Analysis of Option (C): If the expansion is carried out isothermally, the work done is given by $w = -nRT_1 \ln \left( \frac{P_1}{P_2} \right)$.

1. Process: Irreversible isothermal expansion.
2. Work done against constant external pressure: $w = -P_{ext}\Delta V = -P_2(V_2 - V_1)$.
3. Isothermal condition: $T_1 = T_2 = T$.
4. Using Ideal Gas Law: $V_1 = \frac{nRT_1}{P_1}$ and $V_2 = \frac{nRT_1}{P_2}$.
5. Substitute $V_1$ and $V_2$ into the work equation: $w = -P_2\left(\frac{nRT_1}{P_2} - \frac{nRT_1}{P_1}\right)$ $w = -nRT_1\left(1 - \frac{P_2}{P_1}\right) = -nRT_1\left(\frac{P_1 - P_2}{P_1}\right)$ The expression given in the option, $w = -nRT_1 \ln \left( \frac{P_1}{P_2} \right)$, is for a reversible isothermal expansion. This statement is incorrect.

Analysis of Option (D): If the expansion is carried out isothermally, the entropy change is given by $\Delta S = -nR \ln \left( \frac{P_1}{P_2} \right)$.

1. Process: Isothermal expansion. For an ideal gas, $\Delta S_{sys}$ depends only on the initial and final states, regardless of the path (reversible or irreversible).
2. Entropy change for an ideal gas: $\Delta S = nC_v \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)$.
3. Isothermal condition: $T_2 = T_1$, so the first term is zero. $\Delta S = nR \ln\left(\frac{V_2}{V_1}\right)$.
4. Using Ideal Gas Law for isothermal process: $P_1V_1 = P_2V_2 \implies \frac{V_2}{V_1} = \frac{P_1}{P_2}$.
5. Substitute into the entropy equation: $\Delta S = nR \ln\left(\frac{P_1}{P_2}\right)$. The given expression is $\Delta S = -nR \ln \left( \frac{P_1}{P_2} \right) = nR \ln \left( \frac{P_2}{P_1} \right)$. Since it's an expansion, $P_1 > P_2$, so $P_1/P_2 > 1$, and $\ln(P_1/P_2) > 0$. Thus, $\Delta S$ should be positive (entropy increases for expansion). The given expression with the negative sign would result in a negative $\Delta S$, which is incorrect for an expansion. This statement is incorrect.

Final Conclusion: Only statement (A) is correct.

The final answer is $\boxed{A}$

Subject: Chemistry Chapter: Thermodynamics Topic: Work, Heat, Internal Energy, and Entropy Changes in Thermodynamic Processes (specifically Irreversible Processes for Ideal Gases)