A gas expands with temperature according to the relation `V=KT^(2/3)`.Work done when the temperature changes by 60K is.

Options

A. 10R
B. 30R
C. 40R
D. 20R

Explanation

To solve the problem of calculating the work done when a gas expands according to the relation \( V = K T^{2/3} \) and the temperature changes by 60 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Relation:**
We are given the volume \( V \) as a function of temperature \( T \):
\[ V = K T^{2/3} \]

2. **Determine the Change in Temperature:**
The change in temperature is given as:
\[ \Delta T = T_2 - T_1 = 60 \, \text{K} \]

3. **Use the Ideal Gas Law:**
From the ideal gas law, we know:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)):
\[ PV = RT \quad \Rightarrow \quad P = \frac{RT}{V} \]

4. **Calculate the Work Done:**
The work done \( W \) during an expansion can be expressed as:
\[ W = \int P \, dV \]
Substituting for \( P \):
\[ W = \int \frac{RT}{V} \, dV \]

5. **Differentiate the Volume Relation:**
To find \( dV \), differentiate the volume relation:
\[ dV = \frac{d}{dT}(K T^{2/3}) \, dT = K \cdot \frac{2}{3} T^{-1/3} \, dT \]

6. **Substitute \( dV \) into the Work Integral:**
Substitute \( V \) and \( dV \) into the work integral:
\[ W = \int \frac{RT}{K T^{2/3}} \cdot K \cdot \frac{2}{3} T^{-1/3} \, dT \]
Simplifying this gives:
\[ W = \int \frac{2R}{3} T^{1 - 2/3 - 1/3} \, dT = \int \frac{2R}{3} \, dT \]

7. **Integrate Over the Temperature Limits:**
Now, integrate from \( T_1 \) to \( T_2 \):
\[ W = \frac{2R}{3} \int_{T_1}^{T_2} dT = \frac{2R}{3} (T_2 - T_1) = \frac{2R}{3} \cdot 60 \]

8. **Calculate the Final Work Done:**
Thus, the work done is:
\[ W = \frac{2R}{3} \cdot 60 = 40R \]

### Final Answer:
The work done when the temperature changes by 60 K is:
\[ W = 40R \]