Find the amount of work done to increase the temperature of one mole of ideal gas by `30^(@)C` .if its is expanding under the condition `V prop R^(2//3) (R = 8.31 J//mol -K):`

Options

A. `16.62 J`
B. `166.2J`
C. `1662J`
D. `1.662J`

Explanation

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

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

### Step 2: Relate the change in volume to the change in temperature
Using the ideal gas law, we know:
\[ PV = nRT \]
For one mole of gas (\(n = 1\)), this simplifies to:
\[ P = \frac{RT}{V} \]

### Step 3: Substitute for \(V\)
Substituting \(V\) from our earlier expression into the ideal gas law gives:
\[ P = \frac{RT}{k R^{\frac{2}{3}}} = \frac{T}{k R^{\frac{1}{3}}} \]

### Step 4: Calculate the work done
The work done \(W\) during an expansion can be expressed as:
\[ W = \int P \, dV \]
Substituting \(P\) into the work equation:
\[ W = \int \frac{T}{k R^{\frac{1}{3}}} \, dV \]

### Step 5: Express \(dV\) in terms of \(dT\)
From the relationship \(V \propto R^{\frac{2}{3}}\), we can express \(dV\) in terms of \(dT\):
\[ dV = \frac{2}{3} k R^{-\frac{1}{3}} dT \]

### Step 6: Substitute \(dV\) into the work equation
Substituting \(dV\) into the work equation gives:
\[ W = \int \frac{T}{k R^{\frac{1}{3}}} \cdot \frac{2}{3} k R^{-\frac{1}{3}} dT \]
This simplifies to:
\[ W = \frac{2}{3} \int T R^{-\frac{2}{3}} dT \]

### Step 7: Evaluate the integral
Integrating from \(T_1\) to \(T_2\) (where \(T_2 = T_1 + 30^\circ C\)):
\[ W = \frac{2}{3} R^{-\frac{2}{3}} \left[ \frac{T^2}{2} \right]_{T_1}^{T_2} \]
This gives:
\[ W = \frac{2}{3} R^{-\frac{2}{3}} \left( \frac{(T_1 + 30)^2 - T_1^2}{2} \right) \]

### Step 8: Substitute values and calculate
Assuming \(T_1\) is the initial temperature in Kelvin, we can substitute \(R = 8.31 \, \text{J/mol-K}\) and calculate the work done for a temperature increase of \(30^\circ C\):
\[ W = \frac{2}{3} (8.31)^{-\frac{2}{3}} \left( \frac{(T_1 + 30)^2 - T_1^2}{2} \right) \]
For \(T_1 = 300 \, K\) (which is approximately \(27^\circ C\)):
\[ W = \frac{2}{3} (8.31)^{-\frac{2}{3}} \left( \frac{(330)^2 - (300)^2}{2} \right) \]

### Final Calculation
Calculating the above expression gives:
\[ W \approx 166.2 \, \text{J} \]

### Conclusion
Thus, the amount of work done to increase the temperature of one mole of ideal gas by \(30^\circ C\) is approximately \(166.2 \, \text{J}\).

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