Question

For an ideal gas undergo isothermal reversible process from 0.5Mpa, 20dm$^3$ to 0.2Mpa at 600K. Calculate correct option. [Given $\log 5 = 0.6989, \log 2 = 0.3010$]

• A. w = -3.9 kJ, $\Delta$U= 0, q = 3.9 kJ
• B. w = -9.1 kJ, $\Delta$U= 0, q = 9.1 kJ
• C. w = +9.1 kJ, $\Delta$U= 0, q = -9.1 kJ
• D. w = +3.9 kJ, $\Delta$U= 0, q = -3.9 kJ

Hint:

Isothermal reversible work: $w = -2.303 PV \log(V_2/V_1)$. Pressure decrease implies volume increase (Expansion), so work is negative.

Answer 1

The Correct Option is B

Ideal gas, Isothermal $\implies \Delta U = 0$.
Expansion (Pressure decreases from 0.5 to 0.2 MPa) $\implies$ Work done by gas $\implies w<0$.
This eliminates options (3) and (4).
Work Formula: $w = -nRT \ln(P_1/P_2) = -P_1 V_1 \ln(P_1/P_2)$.
$P_1 = 0.5 \text{ MPa} = 5 \times 10^5 \text{ Pa}$.
$V_1 = 20 \text{ dm}^3 = 20 \times 10^{-3} \text{ m}^3$.
$P_1 V_1 = (5 \times 10^5)(2 \times 10^{-2}) = 10^4 \text{ J} = 10 \text{ kJ}$.
Log term: $\ln(0.5/0.2) = \ln(2.5) = 2.303 \log(2.5)$.
$\log 2.5 = \log(10/4) = 1 - 2\log 2 = 1 - 0.6020 = 0.398$.
$\ln(2.5) \approx 2.303 \times 0.398 \approx 0.916$.
Calculation: $w = -10 \text{ kJ} \times 0.916 = -9.16 \text{ kJ}$.
$q = -w = +9.16 \text{ kJ}$.