A diamtomic molecule can be modelled as two rigid balls connected with a spring such that the balls can vibrate with respect to the center of mass of the system ( spring+balls) . Consider a diamtomic gas made of such diatomic molecules. If the gas performs 20 J of work under isobaric condition, then heat given to the gas ( in J ) is

Answer

To solve the problem of finding the heat given to a diatomic gas that performs 20 J of work under isobaric conditions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Degrees of Freedom**:
For a diatomic molecule, the degrees of freedom (f) are 5 (3 translational + 2 rotational).

2. **Calculate Specific Heat at Constant Volume (C_v)**:
The specific heat at constant volume for a diatomic gas is given by:
\[ C_v = \frac{fR}{2} = \frac{5R}{2} \]

3. **Calculate Specific Heat at Constant Pressure (C_p)**:
The specific heat at constant pressure is related to \(C_v\) by:
\[ C_p = C_v + R = \frac{5R}{2} + R = \frac{7R}{2} \]

4. **Work Done Under Isobaric Conditions**:
We know that the work done (W) under isobaric conditions is given as 20 J.

5. **Relate Heat Transfer (Q) to Work Done (W)**:
Under isobaric conditions, the heat added to the system can be expressed as:
\[ Q = nC_p\Delta T \]
where \(\Delta T\) is the change in temperature.

6. **Express \(\Delta T\) in Terms of Work Done**:
From the ideal gas law, we know that:
\[ W = P\Delta V = nR\Delta T \]
Rearranging gives:
\[ \Delta T = \frac{W}{nR} \]

7. **Substitute \(\Delta T\) into the Heat Equation**:
Substitute \(\Delta T\) back into the heat equation:
\[ Q = nC_p\left(\frac{W}{nR}\right) = C_p\frac{W}{R} \]

8. **Substitute \(C_p\)**:
Now substitute \(C_p = \frac{7R}{2}\):
\[ Q = \frac{7R}{2} \cdot \frac{W}{R} = \frac{7}{2}W \]

9. **Substitute the Given Work Done**:
Given \(W = 20 J\):
\[ Q = \frac{7}{2} \cdot 20 = 70 J \]

### Final Answer:
The heat given to the gas is **70 J**.

---