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 solve the problem of finding the work done by one mole of an ideal gas that expands slowly according to the law \( \frac{p}{V} = \text{constant} \), we can follow these steps:

### Step 1: Understand the given relationship
The relationship \( \frac{p}{V} = k \) (where \( k \) is a constant) implies that \( p = kV \). This means that the pressure \( p \) is directly proportional to the volume \( V \).

### Step 2: Set up the work done formula
The work done \( W \) by the gas during an expansion is given by the integral:
\[ W = \int_{V_1}^{V_2} p \, dV \]
Substituting \( p \) from the relationship we have:
\[ W = \int_{V_1}^{V_2} kV \, dV \]

### Step 3: Perform the integration
Now, we can integrate:
\[ W = k \int_{V_1}^{V_2} V \, dV = k \left[ \frac{V^2}{2} \right]_{V_1}^{V_2} \]
Calculating the limits:
\[ W = k \left( \frac{V_2^2}{2} - \frac{V_1^2}{2} \right) = \frac{k}{2} (V_2^2 - V_1^2) \]

### Step 4: Substitute for \( k \)
From the initial condition, we know:
\[ k = \frac{p_1}{V_1} = \frac{p_2}{V_2} \]
Thus, we can express \( k \) in terms of pressure and volume:
\[ W = \frac{p_2}{V_2} \cdot \frac{1}{2} (V_2^2 - V_1^2) \]

### Step 5: Use the ideal gas law
According to the ideal gas law:
\[ PV = nRT \]
For one mole of gas, we have:
\[ p_1V_1 = RT_1 \quad \text{and} \quad p_2V_2 = RT_2 \]
Substituting these into our work equation:
\[ W = \frac{1}{2} \left( \frac{RT_2}{V_2} V_2^2 - \frac{RT_1}{V_1} V_1^2 \right) \]
This simplifies to:
\[ W = \frac{R}{2} (T_2 - T_1) \]

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