The volume of one mole of an ideal gas with adiabatic exponent $\gamma$ is varied according to the l

Question

The volume of one mole of an ideal gas with adiabatic exponent $\gamma$ is varied according to the law $V = a/T$, where $a$ is a constant. What is the amount of heat obtained by the gas in this process if the gas temperature is increased by $\Delta T$?

Professor PC Thomas Classes · Answer

Given the process equation $V = a/T$, we can rewrite this as $VT = a$. Using the ideal gas equation for 1 mole, we have $T = \frac{PV}{R}$. Substituting this expression for $T$ into the process equation yields $V\left(\frac{PV}{R}ight) = a \implies PV^2 = aR = 	ext{constant}$. This is a polytropic process of the general form $PV^n = 	ext{constant}$, where the polytropic index is $n = 2$. The molar heat capacity $C$ for a polytropic process is given by $C = C_v + \frac{R}{1-n}$. We know that $C_v = \frac{R}{\gamma-1}$. Substituting $n=2$, we obtain the molar heat capacity for this specific process: $C = \frac{R}{\gamma-1} + \frac{R}{1-2} = \frac{R}{\gamma-1} - R = R \left( \frac{1 - (\gamma - 1)}{\gamma - 1} ight) = R \frac{2-\gamma}{\gamma-1}$. The heat obtained by the gas is given by $\Delta Q = n_{	ext{moles}} C \Delta T$. Since it is one mole, $\Delta Q = (1) \left( R \frac{2-\gamma}{\gamma-1} ight) \Delta T$.