The pressure of an ideal gas varies with volume as P= aV, where a is a constant. One mole of the gas is allowed to undergo expansion such that its volume becomes ‘m’ times its initial volume. The work done by the gas in the process is

Options

A. `(alphaV)/2 (m^(2)-1)`
B. `(alpha^(2)V^(2))/2(m^(2)-1)`
C. `alpha/2(m^(2)-1)`
D. `(alphaV^(2))/2(m^(2)-1)`

Explanation

To find the work done by the gas during the expansion, we start with the given relationship between pressure and volume:

1. **Identify the relationship**: The pressure \( P \) of the gas varies with volume \( V \) as:
\[ P = aV \]
where \( a \) is a constant.

2. **Understand the process**: We have one mole of the gas expanding from an initial volume \( V_1 \) to a final volume \( V_2 \), where \( V_2 = mV_1 \).

3. **Set up the work done formula**: The work done \( W \) by the gas during an expansion is given by the integral:
\[ W = \int_{V_1}^{V_2} P \, dV \]

4. **Substitute the pressure function**: We substitute the expression for pressure into the work done formula:
\[ W = \int_{V_1}^{V_2} aV \, dV \]

5. **Perform the integration**: The integral of \( aV \) with respect to \( V \) is:
\[ W = a \int_{V_1}^{V_2} V \, dV = a \left[ \frac{V^2}{2} \right]_{V_1}^{V_2} \]
Evaluating this from \( V_1 \) to \( V_2 \):
\[ W = a \left( \frac{(mV_1)^2}{2} - \frac{V_1^2}{2} \right) \]
\[ W = a \left( \frac{m^2 V_1^2}{2} - \frac{V_1^2}{2} \right) \]
\[ W = a \left( \frac{V_1^2}{2} (m^2 - 1) \right) \]

6. **Express \( V_1 \) in terms of the ideal gas law**: Since we have one mole of gas, we can use the ideal gas law \( PV = nRT \) to express \( V_1 \):
\[ V_1 = \frac{RT}{P_1} \]
where \( P_1 \) is the initial pressure.

7. **Final expression for work done**: Substituting \( V_1 \) back into the work done expression:
\[ W = a \left( \frac{1}{2} \left( \frac{RT}{P_1} \right)^2 (m^2 - 1) \right) \]

Thus, the work done by the gas during the expansion is:
\[ W = \frac{a R^2 T^2 (m^2 - 1)}{2 P_1^2} \]