Q69 / 180 · NEET 2026 (Re-NEET) · Chemistry
Two moles of an ideal gas undergo free expansion from $10$ L to $100$ L at $300$ K. The values of $\Delta S_{\text{system}}$ and $\Delta S_{\text{surroundings}}$ are
(R is universal gas constant)
(R is universal gas constant)
- Correct answerA$\Delta S_{\text{system}}=4.606$ R; $\Delta S_{\text{surroundings}}=0$
- B$\Delta S_{\text{system}}=0$; $\Delta S_{\text{surroundings}}=0$
- C$\Delta S_{\text{system}}=4.606$ R; $\Delta S_{\text{surroundings}}=-4.606$ R
- D$\Delta S_{\text{system}}=0$; $\Delta S_{\text{surroundings}}=4.606$ R
Solution
For the free expansion of an ideal gas, the external pressure is zero, so the work done $W = 0$.
Since the process is isothermal at $300$ K, the change in internal energy $\Delta U = 0$.
From the first law of thermodynamics, $q = \Delta U - W = 0$.
Since no heat is exchanged with the surroundings, the change in entropy of the surroundings is:
$\Delta S_{\text{surroundings}} = \dfrac{q_{\text{surr}}}{T} = 0$
Entropy is a state function, so the change in entropy of the system for an isothermal expansion is calculated using a reversible path between the same initial and final states:
$\Delta S_{\text{system}} = nR \ln\left(\dfrac{V_2}{V_1}\right)$
Substituting $n = 2$ moles, $V_1 = 10$ L, and $V_2 = 100$ L:
$\Delta S_{\text{system}} = 2R \ln\left(\dfrac{100}{10}\right) = 2R \ln(10)$
Using $\ln(10) = 2.303 \log_{10}(10) = 2.303$:
$\Delta S_{\text{system}} = 2 \times 2.303 R = 4.606 R$
Answer: $\Delta S_{\text{system}}=4.606$ R; $\Delta S_{\text{surroundings}}=0$
Since the process is isothermal at $300$ K, the change in internal energy $\Delta U = 0$.
From the first law of thermodynamics, $q = \Delta U - W = 0$.
Since no heat is exchanged with the surroundings, the change in entropy of the surroundings is:
$\Delta S_{\text{surroundings}} = \dfrac{q_{\text{surr}}}{T} = 0$
Entropy is a state function, so the change in entropy of the system for an isothermal expansion is calculated using a reversible path between the same initial and final states:
$\Delta S_{\text{system}} = nR \ln\left(\dfrac{V_2}{V_1}\right)$
Substituting $n = 2$ moles, $V_1 = 10$ L, and $V_2 = 100$ L:
$\Delta S_{\text{system}} = 2R \ln\left(\dfrac{100}{10}\right) = 2R \ln(10)$
Using $\ln(10) = 2.303 \log_{10}(10) = 2.303$:
$\Delta S_{\text{system}} = 2 \times 2.303 R = 4.606 R$
Answer: $\Delta S_{\text{system}}=4.606$ R; $\Delta S_{\text{surroundings}}=0$
Hence, the correct answer is option: 'A'.