When an ideal gas `(gamma = 5//3)` is heated under constant pressure, what percentage of given heat energy will be utilized in doing external work ?

Options

A. `40%`
B. `30%`
C. `60%`
D. `20%`

Explanation

To solve the problem of determining the percentage of heat energy utilized in doing external work when an ideal gas (with γ = 5/3) is heated under constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First Law of Thermodynamics**:
The first law states that the change in internal energy (ΔU) is equal to the heat added to the system (ΔQ) minus the work done by the system (ΔW):
\[ \Delta Q = \Delta U + \Delta W \]

2. **Rearrange the Equation**:
We can rearrange the equation to express the work done (ΔW) in terms of heat added (ΔQ) and change in internal energy (ΔU):
\[ \Delta W = \Delta Q - \Delta U \]

3. **Express Work Done as a Fraction of Heat Added**:
We want to find the fraction of heat energy that goes into doing work:
\[ \frac{\Delta W}{\Delta Q} = 1 - \frac{\Delta U}{\Delta Q} \]

4. **Determine Expressions for ΔQ and ΔU**:
For an ideal gas, the heat added at constant pressure (ΔQ) is given by:
\[ \Delta Q = N C_p \Delta T \]
where \(C_p\) is the specific heat at constant pressure.

The change in internal energy (ΔU) for an ideal gas is given by:
\[ \Delta U = N C_v \Delta T \]
where \(C_v\) is the specific heat at constant volume.

5. **Substitute ΔQ and ΔU into the Fraction**:
Now, substituting these expressions into our fraction:
\[ \frac{\Delta W}{\Delta Q} = 1 - \frac{N C_v \Delta T}{N C_p \Delta T} \]
Here, \(N\) and \(\Delta T\) cancel out:
\[ \frac{\Delta W}{\Delta Q} = 1 - \frac{C_v}{C_p} \]

6. **Use the Relationship Between Cp and Cv**:
We know that:
\[ \frac{C_p}{C_v} = \gamma \]
Therefore, we can express \(C_v\) in terms of \(C_p\):
\[ C_v = \frac{C_p}{\gamma} \]

7. **Substitute into the Fraction**:
Now substituting \(C_v\) into our fraction:
\[ \frac{\Delta W}{\Delta Q} = 1 - \frac{C_p/\gamma}{C_p} = 1 - \frac{1}{\gamma} \]

8. **Calculate the Value**:
Given that \(\gamma = \frac{5}{3}\):
\[ \frac{\Delta W}{\Delta Q} = 1 - \frac{1}{5/3} = 1 - \frac{3}{5} = \frac{2}{5} \]

9. **Convert to Percentage**:
To find the percentage of heat energy utilized in doing external work:
\[ \text{Percentage} = \left(\frac{\Delta W}{\Delta Q}\right) \times 100 = \frac{2}{5} \times 100 = 40\% \]

### Final Answer:
The percentage of the given heat energy that will be utilized in doing external work is **40%**.