Question

One mole of a monoatomic ideal gas starting from state A, goes through B and C to state D, as shown in the figure. Total magnitude of change in entropy (in J K⁻¹) during this process is ________ (Nearest integer)

Answer 1

9 J/K

Answer 2

We first determine the temperatures (in arbitrary units proportional to PV) at the four states (using $PV\propto T$ for an ideal gas):

• State A: $P=2$ atm, $V=10$ dm³ $\Rightarrow T_A\propto 2\times10=20$.
• State B: $P=2$ atm, $V=20$ dm³ $\Rightarrow T_B\propto 2\times20=40$.
• State C: $P$ drops to 1 atm at constant volume $20$ dm³ $\Rightarrow T_C\propto 1\times20=20$.
• State D: $P=1$ atm, $V=10$ dm³ $\Rightarrow T_D\propto 1\times10=10$.

Step 1. Process AB (Isobaric expansion at $P=2$ atm):

For an isobaric process the entropy change is:

$$\Delta S_{AB} = n C_P \ln \frac{T_B}{T_A}.$$

For a monoatomic gas,

$$C_P = \frac{5}{2} R,\quad R=\frac{25}{3}\, \text{J/mol·K}\quad \Rightarrow\quad C_P = \frac{5}{2}\times\frac{25}{3}=\frac{125}{6}\, \text{J/K}.$$

And

$$\ln\frac{T_B}{T_A} = \ln\frac{40}{20}=\ln 2\approx 0.7.$$

Thus,

$$\Delta S_{AB} = \frac{125}{6}\times 0.7\approx 14.58\, \text{J/K}.$$

Step 2. Process BC (Isochoric at $V=20$ dm³):

For an isochoric process:

$$\Delta S_{BC} = n C_V \ln \frac{T_C}{T_B}.$$

For a monoatomic gas,

$$C_V = \frac{3}{2}R = \frac{3}{2}\times\frac{25}{3}=\frac{25}{2}=12.5\, \text{J/K}.$$

And

$$\ln\frac{T_C}{T_B} = \ln\frac{20}{40}=\ln\left(\frac{1}{2}\right)=-\ln 2 \approx -0.7.$$

Thus,

$$\Delta S_{BC}=12.5\times(-0.7)=-8.75\, \text{J/K}.$$

Step 3. Process CD (Isobaric compression at $P=1$ atm):

Again for an isobaric process:

$$\Delta S_{CD} = n C_P \ln \frac{T_D}{T_C}.$$

Here,

$$\ln\frac{T_D}{T_C}=\ln\frac{10}{20}=\ln\left(\frac{1}{2}\right)=-\ln 2\approx -0.7.$$

So,

$$\Delta S_{CD}=\frac{125}{6}\times (-0.7)\approx -14.58\, \text{J/K}.$$

Total entropy change from A to D:

$$\Delta S_{total} = \Delta S_{AB} + \Delta S_{BC} + \Delta S_{CD} = 14.58 - 8.75 - 14.58 \approx -8.75\, \text{J/K}.$$

Since the question asks for the total magnitude of change in entropy, we take the absolute value:

$$|\Delta S_{total}| \approx 8.75\, \text{J/K}\quad \text{(Nearest integer: }9\text{ J/K)}.$$

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Explanation (minimal):
Calculate temperatures from $PV$ values. Use $\Delta S = n\,C_P\ln\frac{T_2}{T_1}$ for isobaric and $\Delta S = n\,C_V\ln\frac{T_2}{T_1}$ for isochoric processes. Sum the entropy changes along A→B, B→C, and C→D, then take the absolute value.