One mole of an ideal gas at temperature `T_` expands slowly according to the law `p/V=` constant.
Its final temperature is `T_2`. The work done by the gas is

Options

A. `R(T_(2)-T_(1))`
B. `2R(T_(2)-T_(1))`
C. `(R)/(2)(T_(2)-T_(1))`
D. `(2R)/(3)(T_(2)-T_(1))`

Answer

To find the work done by one mole of an ideal gas expanding according to the law \( \frac{P}{V} = \text{constant} \), we can follow these steps:

### Step 1: Define the Constant
Let the constant \( k \) be defined as:
\[ k = \frac{P}{V} \]
From this, we can express pressure \( P \) in terms of volume \( V \):
\[ P = kV \]

### Step 2: Work Done by the Gas
The work done \( W \) by the gas during expansion can be expressed as:
\[ W = \int_{V_1}^{V_2} P \, dV \]
Substituting \( P \) from the previous step:
\[ W = \int_{V_1}^{V_2} kV \, dV \]

### Step 3: Integrate
Taking \( k \) out of the integral, we have:
\[ W = k \int_{V_1}^{V_2} V \, dV \]
The integral of \( V \) is:
\[ \int V \, dV = \frac{V^2}{2} \]
Thus, we can write:
\[ W = k \left[ \frac{V^2}{2} \right]_{V_1}^{V_2} \]
Evaluating this from \( V_1 \) to \( V_2 \):
\[ W = k \left( \frac{V_2^2}{2} - \frac{V_1^2}{2} \right) = \frac{k}{2} (V_2^2 - V_1^2) \]

### Step 4: Substitute the Constant \( k \)
Since \( k = \frac{P}{V} \), we can express \( P \) in terms of \( V \):
Using the ideal gas law \( PV = nRT \), for one mole of gas (\( n = 1 \)):
\[ PV = RT \implies P = \frac{RT}{V} \]
Thus, substituting for \( k \):
\[ k = \frac{RT}{V} \]
Now substituting this into the work equation:
\[ W = \frac{RT}{2} \left( \frac{V_2^2}{V_2} - \frac{V_1^2}{V_1} \right) = \frac{RT}{2} \left( V_2 - V_1 \right) \]

### Step 5: Relate Temperatures
Using the ideal gas law again, we can relate the temperatures:
\[ P_1V_1 = RT_1 \quad \text{and} \quad P_2V_2 = RT_2 \]
Thus, we can express the work done in terms of temperatures:
\[ W = \frac{R}{2} (T_2 - T_1) \]

### Final Result
The work done by the gas is:
\[ W = \frac{R}{2} (T_2 - T_1) \]