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 gaa 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 280 J of heat is supplied to a diatomic gas at constant pressure.

### Step-by-Step Solution:

1. **Understand 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,
- \( W \) is the work done by the system.

2. **Identify the Given Values**:
From the problem, we know:
- Heat supplied, \( Q = 280 \, \text{J} \).

3. **Work Done by the Gas at Constant Pressure**:
For a gas at constant pressure, the work done by the gas can be expressed as:
\[ W = P \Delta V \]
However, we can also relate the work done to the heat supplied using the specific heat at constant pressure \( C_p \):
\[ Q = n C_p \Delta T \]
For a diatomic gas, the molar specific heat at constant pressure \( C_p \) is given by:
\[ C_p = \frac{7}{2} R \]
where \( R \) is the universal gas constant.

4. **Change in Internal Energy for a Diatomic Gas**:
The change in internal energy for a diatomic gas can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where \( C_v \) for a diatomic gas is:
\[ C_v = \frac{5}{2} R \]

5. **Relate \( Q \), \( W \), and \( \Delta U \)**:
From the first law:
\[ Q = \Delta U + W \]
Rearranging gives:
\[ W = Q - \Delta U \]

6. **Expressing \( \Delta U \) in Terms of \( Q \)**:
We can express \( \Delta U \) in terms of \( Q \):
\[ \Delta U = n C_v \Delta T = n \left(\frac{5}{2} R\right) \Delta T \]
Using the relation for \( Q \):
\[ Q = n C_p \Delta T = n \left(\frac{7}{2} R\right) \Delta T \]

7. **Finding the Ratio**:
Now, substituting \( W \) and \( \Delta U \):
\[ W = Q - \Delta U = n \left(\frac{7}{2} R\right) \Delta T - n \left(\frac{5}{2} R\right) \Delta T \]
\[ W = n \left(\frac{2}{2} R\right) \Delta T = n R \Delta T \]
Now, we can find the ratio:
\[ \frac{W}{\Delta U} = \frac{n R \Delta T}{n \left(\frac{5}{2} R\right) \Delta T} = \frac{2}{5} \]

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