For one mole of an ideal monoatomic gas, volume and temperature are related as `V = KT^(2/3)`. If change in temperature of gas is 90K, then work done is given by W = xR. Find value of x

Options

A. 60
B. 20
C. 90
D. 30

Answer

To solve the problem, we need to find the work done by one mole of an ideal monoatomic gas when the volume and temperature are related by the equation \( V = K T^{2/3} \) and the change in temperature is \( \Delta T = 90 \, K \).

### Step-by-step Solution:

1. **Understand the relationship between volume and temperature**:
We are given that \( V = K T^{2/3} \). This means that the volume of the gas is proportional to the temperature raised to the power of \( \frac{2}{3} \).

2. **Use the ideal gas law**:
The ideal gas law states that \( PV = nRT \). For one mole of gas (\( n = 1 \)), this simplifies to:
\[ PV = RT \]

3. **Express pressure (P) in terms of T**:
From the ideal gas law, we can express pressure as:
\[ P = \frac{RT}{V} \]
Substituting the expression for \( V \) from step 1, we get:
\[ P = \frac{RT}{K T^{2/3}} = \frac{R}{K} T^{1/3} \]

4. **Calculate the differential volume (dV)**:
To find \( dV \), we differentiate the volume equation \( V = K T^{2/3} \):
\[ dV = K \cdot \frac{2}{3} T^{-1/3} dT \]

5. **Substitute P and dV into the work done equation**:
The work done \( W \) by the gas during a change in volume is given by:
\[ W = \int P \, dV \]
Substituting \( P \) and \( dV \) gives:
\[ W = \int \left( \frac{R}{K} T^{1/3} \right) \left( K \cdot \frac{2}{3} T^{-1/3} dT \right) \]
Simplifying this, we have:
\[ W = \int \frac{2R}{3} dT \]

6. **Evaluate the integral**:
The integral of \( dT \) over the change in temperature \( \Delta T \) (which is 90 K) is:
\[ W = \frac{2R}{3} \Delta T = \frac{2R}{3} \cdot 90 = 60R \]

7. **Conclusion**:
The work done by the gas is given by \( W = 60R \). Thus, the value of \( x \) is 60.

### Final Answer:
\[ x = 60 \]