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 \( V = K T^{2/3} \) as the temperature changes by 60 K, we can follow these steps:

### Step 1: Understand the relationship between volume and temperature
Given the relation \( V = K T^{2/3} \), we can express the volume \( V \) in terms of temperature \( T \) and a constant \( K \).

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

### Step 3: Use the ideal gas law to find pressure
From the ideal gas law, we know that \( P = \frac{nRT}{V} \). For 1 mole of gas (\( n = 1 \)):
\[ P = \frac{RT}{V} = \frac{RT}{K T^{2/3}} = \frac{3RT^{1/3}}{K} \]

### Step 4: Substitute \( P \) and \( dV \) into the work done formula
The work done \( W \) during the expansion can be expressed as:
\[ W = \int P \, dV \]
Substituting \( P \) and \( dV \):
\[ W = \int \frac{3RT^{1/3}}{K} \cdot \left(K \cdot \frac{2}{3} T^{-1/3} dT\right) \]
This simplifies to:
\[ W = \int 2R dT \]

### Step 5: Evaluate the integral over the temperature change
Now, we evaluate the integral from \( T_1 \) to \( T_2 \) where \( \Delta T = T_2 - T_1 = 60 \, K \):
\[ W = 2R \int_{T_1}^{T_2} dT = 2R [T]_{T_1}^{T_2} = 2R (T_2 - T_1) = 2R \cdot 60 \]
Thus, we get:
\[ W = 120R \]

### Step 6: Final expression for work done
The work done when the temperature changes by 60 K is:
\[ W = 120R \]