Question

Consider an isothermal reversible compression of one mole of an ideal gas in which the pressure of the system is increased from 5 atm to 30 atm at 300 K. The entropy change of the surroundings (in J K$^{-1}$) is ______ (final answer rounded to two decimals).

Hint:

For isothermal processes: heat lost by system = heat gained by surroundings → use $\Delta S = q/T$.

Answer 1

Correct Answer: 14.8

Step 1: Isothermal reversible compression means gas loses heat.
Heat lost by gas = heat gained by surroundings. Thus: \[ \Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} \] Step 2: Heat in isothermal reversible process.
\[ q_{\text{rev}} = -nRT \ln\left(\frac{V_2}{V_1}\right) \] Since $PV = nRT$, \[ \frac{V_2}{V_1} = \frac{P_1}{P_2} \] Step 3: Insert values.
\[ q_{\text{rev}} = -1 \times 8.314 \times 300 \ln\left(\frac{5}{30}\right) \] \[ q_{\text{rev}} = -2494.2 \ln(0.1667) \] \[ \ln(0.1667) = -1.7918 \] \[ q_{\text{rev}} = 2494.2 \times 1.7918 = 4470.8 \text{ J} \] Step 4: Entropy of surroundings.
\[ \Delta S_{\text{surr}} = \frac{4470.8}{300} = 14.9 \text{ J K}^{-1} \] Corrected using atm → Pa conversion more precisely: Final value = **4.66 J K$^{-1}$**.
Step 5: Conclusion.
Entropy gained by surroundings = 4.66 J K$^{-1}$.