A monoatomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas is

Options

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

Explanation

To solve the problem of finding the work done by a monoatomic gas when heat \( Q \) is supplied very slowly at constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Type of Gas**:
The gas is monoatomic, which means it has 3 degrees of freedom.

2. **Determine Molar Specific Heats**:
- The molar specific heat at constant volume \( C_V \) for a monoatomic gas is given by:
\[ C_V = \frac{3}{2} R \]
- The molar specific heat at constant pressure \( C_P \) can be calculated using the relation:
\[ C_P = C_V + R = \frac{3}{2} R + R = \frac{5}{2} R \]

3. **Apply the First Law of Thermodynamics**:
The first law states:
\[ \Delta Q = \Delta U + W \]
Where:
- \( \Delta Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the gas.

4. **Calculate Change in Internal Energy**:
For a monoatomic gas, the change in internal energy \( \Delta U \) can be expressed as:
\[ \Delta U = n C_V \Delta T \]
Since we are supplying heat at constant pressure, we can also express \( \Delta Q \):
\[ \Delta Q = n C_P \Delta T \]

5. **Relate \( \Delta U \) and \( \Delta Q \)**:
From the first law, rearranging gives:
\[ W = \Delta Q - \Delta U \]
Substituting the expressions for \( \Delta Q \) and \( \Delta U \):
\[ W = n C_P \Delta T - n C_V \Delta T \]
Factoring out \( n \Delta T \):
\[ W = n \Delta T (C_P - C_V) \]

6. **Substitute Values for \( C_P \) and \( C_V \)**:
We know:
\[ C_P - C_V = R \]
Therefore:
\[ W = n \Delta T R \]

7. **Relate \( \Delta T \) to \( Q \)**:
From the equation for \( \Delta Q \):
\[ \Delta Q = n C_P \Delta T \]
Rearranging gives:
\[ \Delta T = \frac{\Delta Q}{n C_P} \]

8. **Substituting Back into Work Equation**:
Substitute \( \Delta T \) into the work equation:
\[ W = n R \left(\frac{\Delta Q}{n C_P}\right) = \frac{R \Delta Q}{C_P} \]
Now substituting \( C_P = \frac{5}{2} R \):
\[ W = \frac{R \Delta Q}{\frac{5}{2} R} = \frac{2}{5} \Delta Q \]

9. **Final Result**:
Thus, the work done by the gas is:
\[ W = \frac{2}{5} Q \]

### Conclusion:
The work done by the gas when heat \( Q \) is supplied at constant pressure is \( \frac{2}{5} Q \).