What is the magnitude of work performed by the gas if its temperature is increased by ? — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Using the ideal gas law $PV = RT$, we substitute $P = aT^\alpha \implies (aT^\alpha)V = RT \implies V = \frac{R}{a}T^{1-\alpha}$. Differentiating gives $dV = \frac{R}{a}(1-\alpha)T^{-\alpha}dT$. Work $W = \int P dV = \int (aT^\alpha) \left[ \frac{R}{a}(1-\alpha)T^{-\alpha} \right] dT = \int R(1-\alpha)dT = R(1-\alpha)\Delta T$.