If the ratio of specific heat of a gas of constant pressure to that at constant volume is `gamma`, the change in internal energy of the mass of gas, when the volume changes from `V` to `2V` at constant pressure `p` is

Options

A. `(R)/((gamma - 1))`
B. `pV`
C. `(pV)/((gamma - 1))`
D. `(gamma pV)/((gamma - 1))`

Answer

To solve the problem, we need to find the change in internal energy (ΔU) of a gas when its volume changes from V to 2V at constant pressure (p). Given that the ratio of specific heat at constant pressure (Cp) to that at constant volume (Cv) is represented by γ (gamma), we can use the following relationships:

1. **Understanding the relationship between internal energy and temperature:**
The change in internal energy (ΔU) for an ideal gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \(n\) is the number of moles, \(C_v\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature.

2. **Finding the change in temperature (ΔT):**
Since the process occurs at constant pressure, we can use the ideal gas law:
\[ PV = nRT \]
For the initial state (volume = V):
\[ pV = nRT_1 \quad \text{(1)} \]
For the final state (volume = 2V):
\[ p(2V) = nRT_2 \quad \text{(2)} \]
From equation (1), we can express \(T_1\):
\[ T_1 = \frac{pV}{nR} \]
From equation (2), we can express \(T_2\):
\[ T_2 = \frac{p(2V)}{nR} = \frac{2pV}{nR} \]
Now, we can find the change in temperature:
\[ \Delta T = T_2 - T_1 = \frac{2pV}{nR} - \frac{pV}{nR} = \frac{pV}{nR} \]

3. **Substituting ΔT into the equation for ΔU:**
Now we substitute ΔT back into the equation for ΔU:
\[ \Delta U = n C_v \Delta T = n C_v \left(\frac{pV}{nR}\right) \]
Simplifying this gives:
\[ \Delta U = C_v \frac{pV}{R} \]

4. **Relating Cv to γ:**
We know that:
\[ \gamma = \frac{C_p}{C_v} \]
and also:
\[ C_p = C_v + R \]
Therefore:
\[ C_v = \frac{R}{\gamma - 1} \]

5. **Substituting Cv into the ΔU equation:**
Now substituting \(C_v\) into the ΔU equation:
\[ \Delta U = \left(\frac{R}{\gamma - 1}\right) \frac{pV}{R} \]
This simplifies to:
\[ \Delta U = \frac{pV}{\gamma - 1} \]

Thus, the change in internal energy of the gas when the volume changes from V to 2V at constant pressure p is:
\[ \Delta U = \frac{pV}{\gamma - 1} \]