Match Expansion Processes of Ideal Gas with Thermodynamic Changes

To match the expansion processes of the ideal gas in Column I with the thermodynamic changes in Column II, we analyze each case using the First Law of Thermodynamics, $Q = \Delta U + W$, and the ideal gas equation $PV = nRT$.

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Process (A)

• Description: An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and Chamber II has a vacuum. The valve is opened.
• Analysis:
  • Since the container is insulated, no heat enters or leaves the system: $Q = 0$.
  • Expansion into a vacuum (free expansion) involves no opposing force, so work done is zero: $W = 0$.
  • According to the First Law of Thermodynamics:
    $\Delta U = Q - W = 0$
  • For an ideal gas, internal energy depends solely on temperature ($U = n C_v T$). Since $\Delta U = 0$, the temperature of the gas remains constant ($\Delta T = 0$).
• Matching Options: q

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Process (B)

• Description: An ideal monoatomic gas expands to twice its original volume such that $P \propto \frac{1}{V^2}$.
• Analysis:
  • The process is polytropic with $PV^2 = \text{constant}$ (i.e., exponent $x = 2$).
  • Temperature variation:
    $T = \frac{PV}{nR} \propto \frac{1}{V}$
    Since the gas expands ($V \to 2V$), the temperature decreases ($\Delta T < 0$).
  • Heat calculation:
    The molar heat capacity for a polytropic process $PV^x = \text{constant}$ is given by:
    $C = C_v + \frac{R}{1 - x}$
    For a monoatomic gas, $C_v = \frac{3}{2}R$. Substituting $x = 2$:
    $C = \frac{3}{2}R + \frac{R}{1 - 2} = \frac{3}{2}R - R = \frac{1}{2}R > 0$
    The heat absorbed by the gas is:
    $Q = n C \Delta T$
    Since $C > 0$ and $\Delta T < 0$, we get $Q < 0$, which means the gas loses heat.
• Matching Options: p, r

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Process (C)

• Description: An ideal monoatomic gas expands to twice its original volume such that $P \propto \frac{1}{V^{4/3}}$.
• Analysis:
  • The process is polytropic with $x = \frac{4}{3}$.
  • Temperature variation:
    $T = \frac{PV}{nR} \propto \frac{1}{V^{1/3}}$
    As volume $V$ increases, temperature $T$ decreases ($\Delta T < 0$).
  • Heat calculation:
    The molar heat capacity is:
    $C = C_v + \frac{R}{1 - x} = \frac{3}{2}R + \frac{R}{1 - \frac{4}{3}} = \frac{3}{2}R - 3R = -\frac{3}{2}R$
    The heat absorbed by the gas is:
    $Q = n C \Delta T$
    Since $C < 0$ and $\Delta T < 0$, we have $Q > 0$, which means the gas gains heat.
• Matching Options: p, s

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Process (D)

• Description: An ideal monoatomic gas expands such that $P$ increases as $V$ increases from $V_1$ to $2V_1$, as shown in the graph.
• Analysis:
  • Temperature variation:
    Since both $P$ and $V$ increase during expansion, $T = \frac{PV}{nR}$ must increase ($\Delta T > 0$).
  • Heat calculation:
    • Change in internal energy: $\Delta U = n C_v \Delta T > 0$ (since $\Delta T > 0$).
    • Work done: $W = \int P dV > 0$ (area under the curve in the $P-V$ diagram is positive as volume increases).
    • By the First Law of Thermodynamics:
      $Q = \Delta U + W > 0$
      Thus, the gas gains heat.
• Matching Options: q, s

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Final Matching:

• A $\rightarrow$ q
• B $\rightarrow$ p, r
• C $\rightarrow$ p, s
• D $\rightarrow$ q, s