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Thermodynamics question

2013 · Shift 0 · Q1

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Thermodynamics
2013 · Shift 0 · Q1

Thermodynamics question

2013 · Shift 0 · Q1

JEE MainChemistryThermodynamicsMCQ+4 / −1

A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0oC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be : (R = 8.314 J/mol K) ( l n 7.5 = 2.01)

A
q = – 208 J, w = – 208 J
B
q = – 208 J, w = + 208 J
C
q = + 208 J, w = + 208 J
D
q = + 208 J, w = – 208 J

View written solutionFree

Correct answer: D

Given data

Number of moles: $n = 0.04\ \text{mol}$
Initial volume: $V_1 = 50.0\ \text{mL}$
Final volume: $V_2 = 375\ \text{mL}$
Temperature: $T = 37.0^\circ\text{C} = 310\ \text{K}$
Heat absorbed: $208\ \text{J}$
Process: reversible isothermal expansion of an ideal gas

Sign of heat $q$

The gas absorbs heat, so by convention: $q = +208\ \text{J}$

Internal energy change

For an ideal gas, internal energy depends only on temperature. Since the process is isothermal: $\Delta U = 0$

Using the first law: $\Delta U = q + w$ So, $0 = q + w$ $w = -q = -208\ \text{J}$

Check using reversible isothermal work formula

For reversible isothermal expansion: $w = -nRT \ln\frac{V_2}{V_1}$

Now, $\frac{V_2}{V_1} = \frac{375}{50} = 7.5$

Thus, $w = -(0.04)(8.314)(310)(\ln 7.5)$ Given $\ln 7.5 = 2.01$ , $w = -(0.04)(8.314)(310)(2.01)$

First calculate: $8.314 \times 310 = 2577.34$ $0.04 \times 2577.34 = 103.0936$ $103.0936 \times 2.01 \approx 207.22$

So, $w \approx -207\ \text{J} \approx -208\ \text{J}$

This confirms the result.

Evaluate options

A: $q=-208\ \text{J},\ w=-208\ \text{J}$ ❌
B: $q=-208\ \text{J},\ w=+208\ \text{J}$ ❌
C: $q=+208\ \text{J},\ w=+208\ \text{J}$ ❌
D: $q=+208\ \text{J},\ w=-208\ \text{J}$ ✅

Therefore, the correct option is D.

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