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 \(V \propto T^{2/3}\), we can follow these steps:

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

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

### Step 3: Use the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
Rearranging gives us:
\[ P = \frac{nRT}{V} \]
For one mole of gas (\(n = 1\)), this simplifies to:
\[ P = \frac{RT}{V} \]

### Step 4: Substitute pressure into the work done formula
The work done \(W\) during an expansion is given by:
\[ W = \int P \, dV \]
Substituting the expression for pressure:
\[ W = \int \frac{RT}{V} \, dV \]

### Step 5: Substitute \(dV\) into the work done formula
Now we substitute \(dV\) from Step 2 into the work done formula:
\[ W = \int \frac{RT}{k T^{2/3}} \cdot \frac{2}{3} k T^{-1/3} dT \]
This simplifies to:
\[ W = \frac{2R}{3} \int T^{1/3} dT \]

### Step 6: Evaluate the integral
The integral of \(T^{1/3}\) is:
\[ \int T^{1/3} dT = \frac{3}{4} T^{4/3} \]
Thus, we have:
\[ W = \frac{2R}{3} \cdot \frac{3}{4} \left[ T^{4/3} \right]_{T_1}^{T_2} \]
This simplifies to:
\[ W = \frac{R}{2} \left[ T_2^{4/3} - T_1^{4/3} \right] \]

### Step 7: Calculate the temperature change
The temperature change is given as \(30^\circ C\). We need to convert this to Kelvin:
\[ \Delta T = 30 + 273.15 = 303.15 \, K \]
Assuming \(T_1\) is the initial temperature, we can set \(T_1 = 273.15 \, K\) and \(T_2 = 303.15 \, K\).

### Step 8: Substitute values and calculate work done
Now substituting the values into the work done formula:
\[ W = \frac{8.31}{2} \left[ (303.15)^{4/3} - (273.15)^{4/3} \right] \]

Calculating \(T_2^{4/3}\) and \(T_1^{4/3}\):
\[ W \approx \frac{8.31}{2} \left[ (303.15)^{4/3} - (273.15)^{4/3} \right] \]
After calculating, we find:
\[ W \approx 166.2 \, J \]

### Final Answer
The amount of work done to increase the temperature of one mole of ideal gas by \(30^\circ C\) is approximately \(166.2 \, J\).

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