Question

1 mole gas at 522.27 K performs a reversible work by expansion up to 3.99 times of the Initlal volume. Determine the magnitude work done by gas In kJ. (log 3.99 = 0.6).

Answer 1

6.0 kJ

Answer 2

The work done by a gas during a reversible isothermal expansion is calculated using $W_{by} = nRT \ln(V_f/V_i)$. The common logarithm of the volume ratio was converted to the natural logarithm using $\ln(x) = 2.303 \log_{10}(x)$. Substituting the given values for $n$, $R$, $T$, and $\ln(V_f/V_i)$ yielded the work done in Joules, which was then converted to kilojoules. The calculated value is approximately $6.0$ kJ.