A sample of diatomic gas is heated at constant pressure. If an amount of 280 J of heat is supplied to gas, find ratio of work done by gas and change in internal energy

Options

A. `5:2`
B. `2:5`
C. `2:3`
D. `3:2`

Answer

To solve the problem, we need to find the ratio of work done by the gas (W) to the change in internal energy (ΔU) when a diatomic gas is heated at constant pressure with 280 J of heat supplied.

### Step-by-Step Solution:

1. **Understanding the First Law of Thermodynamics**:
The first law of thermodynamics states that:
\[ \Delta Q = \Delta U + W \]
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 system.

2. **Rearranging the Equation**:
We can rearrange this equation to express work done in terms of heat and change in internal energy:
\[ W = \Delta Q - \Delta U \]

3. **Finding the Ratio**:
We want to find the ratio \(\frac{W}{\Delta U}\). From the rearranged equation, we can express this as:
\[ \frac{W}{\Delta U} = \frac{\Delta Q - \Delta U}{\Delta U} = \frac{\Delta Q}{\Delta U} - 1 \]

4. **Using the Specific Heat Capacities**:
For a diatomic gas, the heat added at constant pressure can be expressed as:
\[ \Delta Q = n C_p \Delta T \]
and the change in internal energy is given by:
\[ \Delta U = n C_v \Delta T \]
Thus, the ratio becomes:
\[ \frac{\Delta Q}{\Delta U} = \frac{n C_p \Delta T}{n C_v \Delta T} = \frac{C_p}{C_v} \]
The terms \(n\) and \(\Delta T\) cancel out.

5. **Using the Heat Capacity Ratio**:
For a diatomic gas, the ratio of specific heats is:
\[ \frac{C_p}{C_v} = \gamma = \frac{7}{5} \]
Therefore:
\[ \frac{\Delta Q}{\Delta U} = \frac{7}{5} \]

6. **Calculating the Ratio of Work Done to Change in Internal Energy**:
Now substituting this value back into our ratio:
\[ \frac{W}{\Delta U} = \frac{7}{5} - 1 = \frac{7 - 5}{5} = \frac{2}{5} \]

7. **Final Result**:
The ratio of work done by the gas to the change in internal energy is:
\[ \frac{W}{\Delta U} = \frac{2}{5} \]

### Conclusion:
The final answer is:
\[ \text{The ratio of work done by gas to change in internal energy is } 2:5. \]