Find the amount of work done to increase the temperature of one mole of an ideal gas by `30^(@)C`, if it is expanding under condition `VooT^(2//3)`.

Options

A. `32R`
B. `15R`
C. `20R`
D. `25R`

Explanation

To find the amount of work done to increase the temperature of one mole of an ideal gas by \(30^\circ C\) while expanding under the condition \(V \propto T^{2/3}\), we can follow these steps:

### Step 1: Understand the relationship between volume and temperature
Given that \(V \propto T^{2/3}\), we can express this as:
\[ V = K T^{2/3} \]
where \(K\) is a constant.

### Step 2: Use the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
For one mole of gas (\(n = 1\)), this simplifies to:
\[ P = \frac{RT}{V} \]
Substituting \(V = K T^{2/3}\) into the equation gives:
\[ P = \frac{RT}{K T^{2/3}} = \frac{R}{K} T^{1/3} \]

### Step 3: Differentiate the volume with respect to temperature
To find \(dV\), we differentiate \(V = K T^{2/3}\):
\[ dV = K \cdot \frac{2}{3} T^{-1/3} dT \]

### Step 4: Write the expression for work done
The work done \(dW\) during a quasi-static process is given by:
\[ dW = P dV \]
Substituting the expressions for \(P\) and \(dV\):
\[ dW = \left(\frac{R}{K} T^{1/3}\right) \left(K \cdot \frac{2}{3} T^{-1/3} dT\right) \]
This simplifies to:
\[ dW = \frac{2R}{3} dT \]

### Step 5: Integrate to find total work done
To find the total work done as the temperature changes from \(T_1\) to \(T_2\), we integrate:
\[ W = \int_{T_1}^{T_2} \frac{2R}{3} dT = \frac{2R}{3} (T_2 - T_1) \]
Given that the change in temperature \(\Delta T = T_2 - T_1 = 30^\circ C\):
\[ W = \frac{2R}{3} \cdot 30 = 20R \]

### Final Answer
Thus, the amount of work done to increase the temperature of one mole of an ideal gas by \(30^\circ C\) is:
\[ W = 20R \]