Question

$1d{m}^3$ of an ideal gas at a pressure of 10 bar expands reversibly and isothermally to final volume of 20 litre. What will be the magnitude of heat involved in the process (in kJ)? (Take ln 20 = 3)

Answer 1

$\Delta U=0\Rightarrow q=-w$

$w=-nRTln\frac{V_2}{V_1}=-P_1{V}_1$ ln $\frac{V_2}{V_1}$

$=-10\times 1ln\frac{20}{1}$

$=-10\times 3$ bar - lit

$=-3\times {10}^5\times {10}^{-3}=-300J$
Hence, the required answer is 0.3 kJ.