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, we need to find the heat given to a diatomic gas that performs 20 J of work under isobaric conditions.

### Step-by-Step Solution:

1. **Understanding the System**:
We have a diatomic gas that can be modeled as two rigid balls connected by a spring. The diatomic gas has 5 degrees of freedom (3 translational and 2 rotational).

2. **Specific Heat Capacities**:
- The specific heat at constant volume (Cv) for a diatomic gas is given by:
\[ C_v = \frac{fR}{2} = \frac{5R}{2} \]
- The specific heat at constant pressure (Cp) can be calculated using the relation:
\[ C_p = C_v + R = \frac{5R}{2} + R = \frac{7R}{2} \]

3. **Work Done by the Gas**:
The work done by the gas (W) is given as 20 J.

4. **Using the First Law of Thermodynamics**:
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.

5. **Change in Internal Energy**:
For an ideal gas, the change in internal energy (\(\Delta U\)) at constant volume can be expressed as:
\[ \Delta U = N C_v \Delta T \]
However, we will express \(\Delta T\) in terms of work done since we are dealing with an isobaric process.

6. **Relating Work and Temperature Change**:
Under isobaric conditions, the work done can also be expressed as:
\[ W = P \Delta V = N R \Delta T \]
Therefore, we can express \(\Delta T\) as:
\[ \Delta T = \frac{W}{N R} \]

7. **Substituting into the First Law**:
Substitute \(\Delta T\) into the equation for \(\Delta U\):
\[ \Delta U = N C_v \left(\frac{W}{N R}\right) = C_v \frac{W}{R} \]
Now substituting this into the first law:
\[ Q = C_v \frac{W}{R} + W \]

8. **Calculating Heat (Q)**:
Substitute \(C_v = \frac{5R}{2}\) and \(W = 20 J\):
\[ Q = \frac{5R}{2} \cdot \frac{20}{R} + 20 \]
Simplifying:
\[ Q = 50 + 20 = 70 J \]

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