The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3ml^-1Jmol^-1K^-1)`

Options

A. diatomic
B. triatomic
C. a mixture of monatomic and diatomic
D. monatomic

Explanation

To solve the problem, we will follow these steps:

1. **Identify the Given Data**:
- Work done (W) = 146 kJ = 146,000 J (since 1 kJ = 1000 J)
- Number of moles (n) = 1 kmol = 1000 mol
- Temperature change (ΔT) = 7°C = 7 K (temperature change is the same in Celsius and Kelvin)
- Universal gas constant (R) = 8.3 J/(mol·K)

2. **Understand the Adiabatic Process**:
- For an adiabatic process, the work done on the gas is related to the change in temperature and the specific heat ratio (γ).
- The formula for work done in an adiabatic process is:
\[ W = nR(T_2 - T_1) / (1 - \gamma) \]
- Here, \(T_2 - T_1 = \Delta T\), which is 7 K.

3. **Rearranging the Formula**:
- We can rearrange the formula to find \(1 - \gamma\):
\[ 1 - \gamma = \frac{nR\Delta T}{W} \]

4. **Substituting the Values**:
- Substitute the known values into the equation:
\[ 1 - \gamma = \frac{(1000 \, \text{mol})(8.3 \, \text{J/(mol·K)})(7 \, \text{K})}{-146000 \, \text{J}} \]

5. **Calculating the Numerator**:
- Calculate the numerator:
\[ 1000 \times 8.3 \times 7 = 58100 \, \text{J} \]

6. **Calculating \(1 - \gamma\)**:
- Now substitute this into the equation:
\[ 1 - \gamma = \frac{58100}{-146000} \]
- This simplifies to:
\[ 1 - \gamma = -0.39726 \approx -0.4 \]

7. **Finding \(\gamma\)**:
- Rearranging gives:
\[ \gamma = 1 + 0.4 = 1.4 \]

8. **Identifying the Gas**:
- The value of \(\gamma = 1.4\) corresponds to a diatomic gas (like O₂, N₂, etc.).

### Final Answer:
The gas is a **diatomic gas**.