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. Triatomic
B. A mixture of monoatomic and idatomic
C. Monoatomic
D. Diatomic

Explanation

To solve the problem step by step, we will use the information provided about the adiabatic process and the properties of the gas.

### Step 1: Understand the given data
We have:
- Work done on the gas, \( W = 146 \, \text{kJ} = 146 \times 10^3 \, \text{J} \)
- Number of moles, \( N = 1 \, \text{kilo mole} = 1000 \, \text{mol} \)
- Temperature change, \( \Delta T = T_2 - T_1 = 7 \, \text{°C} = 7 \, \text{K} \)
- Universal gas constant, \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \)

### Step 2: Apply the formula for work done in an adiabatic process
The work done in an adiabatic process can be expressed as:
\[ W = N R (T_1 - T_2) \frac{1}{\gamma - 1} \]
Since \( T_1 - T_2 = -\Delta T \), we can rewrite it as:
\[ W = -N R \Delta T \frac{1}{\gamma - 1} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ -146 \times 10^3 = -1000 \times 8.3 \times 7 \frac{1}{\gamma - 1} \]

### Step 4: Simplify the equation
Removing the negative signs and rearranging gives:
\[ 146 \times 10^3 = 1000 \times 8.3 \times 7 \frac{1}{\gamma - 1} \]

### Step 5: Calculate the right-hand side
Calculate \( 1000 \times 8.3 \times 7 \):
\[ 1000 \times 8.3 \times 7 = 58100 \]
Thus, we have:
\[ 146 \times 10^3 = \frac{58100}{\gamma - 1} \]

### Step 6: Rearranging for \( \gamma - 1 \)
Rearranging gives:
\[ \gamma - 1 = \frac{58100}{146 \times 10^3} \]

### Step 7: Calculate \( \gamma - 1 \)
Calculating the right-hand side:
\[ \gamma - 1 = \frac{58100}{146000} \approx 0.397 \]

### Step 8: Calculate \( \gamma \)
Adding 1 to both sides:
\[ \gamma \approx 1 + 0.397 \approx 1.397 \]

### Step 9: Identify the type of gas
The value of \( \gamma \approx 1.4 \) indicates that the gas is likely a diatomic gas.

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

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