A gas, for which `gamma` is `(4)/(3)` is heated at constant pressure. The percentage of heat supplied used for external work is

Options

A. `25%`
B. `15%`
C. `60%`
D. `40%`

Explanation

To solve the problem of finding the percentage of heat supplied used for external work when a gas is heated at constant pressure, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- The specific heat ratio (gamma) \( \gamma = \frac{4}{3} \).
- The process is at constant pressure.

### Step 2: Identify Relevant Formulas
1. The work done \( \Delta W \) at constant pressure is given by:
\[ \Delta W = P \Delta V \]
2. The heat supplied \( \Delta Q \) at constant pressure is given by:
\[ \Delta Q = n C_P \Delta T \]
where \( n \) is the number of moles, \( C_P \) is the specific heat at constant pressure, and \( \Delta T \) is the change in temperature.

### Step 3: Relate Work Done to Heat Supplied
Using the ideal gas law, we can express the work done in terms of temperature change:
\[ \Delta W = n R \Delta T \]
where \( R \) is the universal gas constant.

### Step 4: Calculate the Ratio of Work Done to Heat Supplied
Now, we can find the ratio of work done to heat supplied:
\[ \frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_P \Delta T} \]
Here, \( n \) and \( \Delta T \) cancel out:
\[ \frac{\Delta W}{\Delta Q} = \frac{R}{C_P} \]

### Step 5: Express \( R \) in Terms of \( C_P \) and \( \gamma \)
From thermodynamic relations, we know:
\[ C_P = \frac{R}{\gamma - 1} \]
Substituting this into our ratio gives:
\[ \frac{\Delta W}{\Delta Q} = \frac{R}{\frac{R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma} \]

### Step 6: Substitute the Value of \( \gamma \)
Now substituting \( \gamma = \frac{4}{3} \):
\[ \frac{\Delta W}{\Delta Q} = \frac{\frac{4}{3} - 1}{\frac{4}{3}} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4} \]

### Step 7: Convert to Percentage
To find the percentage of heat supplied used for external work:
\[ \text{Percentage} = \left(\frac{\Delta W}{\Delta Q}\right) \times 100 = \frac{1}{4} \times 100 = 25\% \]

### Final Answer
The percentage of heat supplied used for external work is **25%**.

---