Question

A 3.1-mole sample of an ideal gas is gently heated at constant temperature 320 K. It expands from initial volume 23 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).

Answer 1

From the ideal gas law,

PV = nRT ==> P = nRT/V

where P is the pressure exerted by the gas on the container. The work W done by this pressure as the volume of the gas changes from V₁ to V₂ is given by the integral,

`W = \displaystyle \int_(V_1)^(V_2)P\,\mathrm dV \implies W = nRT \ln\left((V_2)/(V_1)\right)`

and solving for V₂ gives

`V_2 = V_1\exp\left((W)/(nRT)\right)`

If you add 1.7 kJ of heat to the system, which does the aforementioned work, the gas will expand to a volume of

`V_2 = (23\,\mathrm L)\exp\left(\frac{1.7\,\mathrm{kJ}}{(3.1\,\mathrm{mol})\left(8.314\frac{\rm J}{\mathrm{mol}\cdot\mathrm K}\right)(320\,\mathrm K)}\right) \approx \boxed{28 \,\mathrm L}`