Question

If the ratio of specific heat of a gas at Constant pressure to that at constant volume is \(\gamma\), the Change in internal energy of the mass of gas, when the volume changes from \(\mathrm{V}\) to \(2 \mathrm{~V}\) at Constant Pressure p, is (A) \(\\{(\mathrm{PV}) /(\gamma-1)\\}\) (B) \(\\{\mathrm{R} /(\gamma-1)\\}\) (C) PV (D) \(\\{(\gamma \mathrm{PV}) /(\gamma-1)\\}\)

Short answer

The change in internal energy of the mass of gas when the volume changes from V to 2V at constant pressure p is given by (D) \(\frac{\gamma PV}{\gamma-1}\).

Step-by-step solution

Write down the given information

The ratio of specific heat at constant pressure (Cp) to that at constant volume (Cv) is given by \(\gamma = \frac{C_{p}}{C_{v}}\). The volume changes from V to 2V under constant pressure p.

Ideal Gas Law

Via the ideal gas law, we know that \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since we have constant pressure, the relationship between the initial and final states is: \[ \frac{P(2V)}{PV}=\frac{T_{2}}{T_{1}} \implies T_{2} = 2T_{1} \]

Specific Heats and Internal Energy relationship

Recall the definition of specific heats at constant volume and constant pressure, \(C_{v}=\frac{dQ}{dT}\), \(C_{p}=\frac{dQ}{dT}\), where dQ is heat transferred, and dT is the change in temperature. Using the information given about the specific heat ratio, which is \(\gamma=\frac{C_{p}}{C_{v}}\), Combine these to obtain: \[ C_{v} =\frac{1}{\gamma-1} C_{p} \] Also, recall the relationship between internal energy (\(dU\)), heat transferred (\(dQ\)), and work done (\(dW\)): \(dU=dQ-dW\). When the process is carried out at constant pressure, work done is given by, \[ dW = PdV \]

Substitute the expressions and solve for internal energy change

Now, the change in internal energy is given by: \[ \Delta U=\int_{T_{1}}^{T_{2}} C_{v}dT \] Substitute the expressions from step 3 for \(C_{v}\) in terms of \(C_{p}\) and step 2 for initial and final temperatures: \[ \Delta U = \frac{1}{\gamma-1}\int_{T_{1}}^{2T_{1}} C_{p} dT \] Since \(C_{p}\) is constant, it can be taken out of the integral: \[ \Delta U = \frac{1}{\gamma-1} C_{p} \int_{T_{1}}^{2T_{1}} dT \] Evaluate the integral: \[ \Delta U=\frac{1}{\gamma-1} C_{p} (2T_{1}-T_{1})= \frac{1}{\gamma-1} C_{p} T_{1} \] Now, substitute the ideal gas law expression for the initial state, \(PV=nRT_{1}\): \[ \Delta U = \frac{1}{\gamma-1}\frac{C_{p}}{R} PV \] Finally, since \(C_{p}=\gamma C_{v}\), and since \(C_{p}-C_{v}=R\), we have \(C_{p}=\gamma R\). Substitute this into the previous equation to obtain the final answer: \[ \Delta U = \frac{\gamma PV}{\gamma-1} \] Thus, the correct answer is (D) \(\frac{\gamma PV}{\gamma-1}\).