$$20.0 dm^{3}$$ of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressme of the gas is 0.2 MPa. Which of the following option is correct?
(Given: $$\log 2 = 0.30 10 and \log 5 = 0.6989$$)

For isothermal reversible expansion of ideal gas: $$\Delta U = 0$$, $$\Delta H = 0$$.

$$w = -nRT\ln\frac{P_1}{P_2}$$. Here $$PV = nRT$$, so $$nRT = 0.5 \times 10^6 \times 20 \times 10^{-3} = 10000$$ J = 10 kJ.

$$w = -10000 \times \ln(0.5/0.2) = -10000 \times \ln 2.5 = -10000 \times 2.303 \times \log 2.5$$

$$= -10000 \times 2.303 \times (0.6989 - 0.3010) = -10000 \times 2.303 \times 0.3979$$

Hmm, $$\log(P_1/P_2) = \log(0.5/0.2) = \log 2.5 = \log(5/2) = 0.6989 - 0.3010 = 0.3979$$

$$w = -2.303 \times 10000 \times 0.3979 = -9163$$ J ≈ -9.1 kJ

$$q = -w = 9.1$$ kJ

The answer is Option 4: w = -9.1 kJ, ΔU = 0, ΔH = 0, q = 9.1 kJ.