A gas expands with temperature according to the relation `V = kT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?

Options

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

Answer

To find the work done when the temperature of a gas changes by \(30^\circ C\) and the volume expands according to the relation \(V = kT^{2/3}\), we can follow these steps:

### Step 1: Understand the Relation
The volume \(V\) of the gas is given by the equation:
\[ V = kT^{2/3} \]
where \(k\) is a constant and \(T\) is the temperature in Kelvin.

### Step 2: Differentiate the Volume
To find the change in volume \(dV\) with respect to temperature \(dT\), we differentiate the volume equation:
\[ dV = \frac{d}{dT}(kT^{2/3}) = k \cdot \frac{2}{3} T^{-1/3} dT \]

### Step 3: Use the Ideal Gas Law
From the ideal gas law, we know:
\[ PV = nRT \]
We can express pressure \(P\) in terms of \(T\) and \(V\):
\[ P = \frac{nRT}{V} \]
Substituting \(V = kT^{2/3}\) into the equation gives:
\[ P = \frac{nRT}{kT^{2/3}} = \frac{nR}{k} T^{1/3} \]

### Step 4: Set Up the Work Integral
The work done \(W\) during the expansion can be expressed as:
\[ W = \int P \, dV \]
Substituting the expressions for \(P\) and \(dV\):
\[ W = \int \left(\frac{nR}{k} T^{1/3}\right) \left(k \cdot \frac{2}{3} T^{-1/3} dT\right) \]
This simplifies to:
\[ W = \int \frac{nR}{k} \cdot k \cdot \frac{2}{3} T^{1/3} T^{-1/3} dT = \int \frac{2nR}{3} dT \]

### Step 5: Evaluate the Integral
The integral becomes:
\[ W = \frac{2nR}{3} \int dT = \frac{2nR}{3} (T_2 - T_1) \]
Given that the temperature change is \(30^\circ C\) (or \(30 K\)):
\[ W = \frac{2nR}{3} \cdot 30 = 20nR \]

### Step 6: Final Result
Assuming we have 1 mole of gas (\(n = 1\)):
\[ W = 20R \]

### Conclusion
Thus, the work done when the temperature changes by \(30^\circ C\) is:
\[ W = 20R \]