Find the amount of work done to increase the temperature of one mole jof an ideal gas by `30^(@)C`, if it is expanding under condition `VooT^(2//3)`.

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-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
Since we have 1 mole of gas, we can simplify this to:
\[ PV = RT \]

2. **Express Volume in Terms of Temperature**:
Given the condition \(V \propto T^{2/3}\), we can express volume as:
\[ V = kT^{2/3} \]
where \(k\) is a constant.

3. **Substitute Volume into the Ideal Gas Law**:
Substitute \(V\) into the ideal gas law:
\[ P(kT^{2/3}) = RT \]
Rearranging gives:
\[ P = \frac{RT}{kT^{2/3}} = \frac{R}{k}T^{1/3} \]
Let \(C = \frac{R}{k}\), then:
\[ P = CT^{1/3} \]

4. **Calculate Work Done**:
The work done \(W\) during an expansion is given by:
\[ W = \int P \, dV \]
We need to express \(dV\) in terms of \(dT\). From \(V = kT^{2/3}\):
\[ dV = k \cdot \frac{2}{3} T^{-1/3} dT \]

5. **Substitute \(P\) and \(dV\) into the Work Equation**:
Substitute \(P\) and \(dV\) into the work formula:
\[ W = \int P \, dV = \int CT^{1/3} \left(k \cdot \frac{2}{3} T^{-1/3} dT\right) \]
This simplifies to:
\[ W = \int \frac{2Ck}{3} dT \]

6. **Integrate**:
The integral becomes:
\[ W = \frac{2Ck}{3} (T_2 - T_1) \]
where \(T_2\) is the final temperature and \(T_1\) is the initial temperature.

7. **Calculate Temperature Change**:
The temperature change is given as \(30^\circ C\), which is equivalent to \(30 K\) (since the change in temperature is the same in both Celsius and Kelvin scales). Thus:
\[ W = \frac{2Ck}{3} \cdot 30 \]

8. **Substitute Constants**:
Recall that \(C = \frac{R}{k}\), so:
\[ W = \frac{2R}{3} \cdot 30 \]
Now substituting \(R = 8.314 \, \text{J/mol K}\):
\[ W = \frac{2 \cdot 8.314}{3} \cdot 30 \]

9. **Calculate Final Work Done**:
\[ W = \frac{16.628}{3} \cdot 30 = 166.28 \, \text{J} \]

### Final Answer:
The amount of work done to increase the temperature of one mole of an ideal gas by \(30^\circ C\) is approximately \(166.28 \, \text{J}\).