A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is

Options

A. Q
B. `(3)/(5) Q`
C. `(2)/(5) Q`
D. `(2)/(3) Q`

Explanation

To solve the problem of finding the work done during the expansion of an ideal monoatomic gas at constant pressure when heat \( Q \) is given to it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between heat, internal energy, and work done**:
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
where \( Q \) is the heat added to the system, \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the system.

2. **Identify the specific heat capacities**:
For a monoatomic ideal gas, the specific heat at constant volume \( C_V \) and constant pressure \( C_P \) are given by:
\[ C_V = \frac{3}{2} R \quad \text{and} \quad C_P = \frac{5}{2} R \]
where \( R \) is the universal gas constant.

3. **Express the heat added in terms of temperature change**:
The heat added at constant pressure can be expressed as:
\[ Q = n C_P \Delta T \]
Substituting the value of \( C_P \):
\[ Q = n \left(\frac{5}{2} R\right) \Delta T \]

4. **Express the change in internal energy**:
The change in internal energy for the gas can be expressed as:
\[ \Delta U = n C_V \Delta T \]
Substituting the value of \( C_V \):
\[ \Delta U = n \left(\frac{3}{2} R\right) \Delta T \]

5. **Substitute \( Q \) and \( \Delta U \) into the first law equation**:
Now we substitute \( Q \) and \( \Delta U \) into the first law equation:
\[ Q = \Delta U + W \]
This gives:
\[ n \left(\frac{5}{2} R\right) \Delta T = n \left(\frac{3}{2} R\right) \Delta T + W \]

6. **Solve for work done \( W \)**:
Rearranging the equation to solve for \( W \):
\[ W = n \left(\frac{5}{2} R\right) \Delta T - n \left(\frac{3}{2} R\right) \Delta T \]
Simplifying this:
\[ W = n R \Delta T \]

7. **Relate \( W \) to \( Q \)**:
From the expression for \( Q \):
\[ Q = n \left(\frac{5}{2} R\right) \Delta T \]
We can express \( \Delta T \) in terms of \( Q \):
\[ \Delta T = \frac{2Q}{5nR} \]
Substituting this back into the expression for \( W \):
\[ W = n R \left(\frac{2Q}{5nR}\right) = \frac{2Q}{5} \]

### Final Answer:
Thus, the work done \( W \) in the expansion is:
\[ W = \frac{2}{5} Q \]