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

Options

A. diatomic
B. triatomic
C. mixture of monoatomic and diatomic
D. monoatomic

Explanation

To solve the problem step by step, we will use the adiabatic process formula and the given values to determine the nature of the gas.

### Step 1: Understand the given data
- Work done (W) = 146 kJ = 146,000 J (since 1 kJ = 1000 J)
- Number of moles (n) = 1 kmole = 1000 moles
- Change in temperature (ΔT) = 7°C = 7 K (since the change in temperature is the same in Celsius and Kelvin)
- Gas constant (R) = 8.3 J/(mol·K)

### Step 2: Use the adiabatic work done formula
In an adiabatic process, the work done on the gas can be expressed as:
\[ W = nR \frac{T_1 - T_2}{\gamma - 1} \]
Where:
- \(T_1\) is the initial temperature,
- \(T_2\) is the final temperature,
- \(\gamma\) is the heat capacity ratio (Cp/Cv).

Given that the temperature increases by 7 K, we can express this as:
\[ T_2 - T_1 = 7 \implies T_1 - T_2 = -7 \]

### Step 3: Substitute the values into the formula
Substituting the known values into the work done formula:
\[ 146,000 = 1000 \times 8.3 \times \frac{-7}{\gamma - 1} \]

### Step 4: Simplify the equation
First, calculate \(1000 \times 8.3\):
\[ 1000 \times 8.3 = 8300 \]
Now substituting back into the equation:
\[ 146,000 = 8300 \times \frac{-7}{\gamma - 1} \]
Rearranging gives:
\[ \frac{-7 \times 8300}{\gamma - 1} = 146,000 \]
\[ -58100 = 146,000(\gamma - 1) \]

### Step 5: Solve for \(\gamma - 1\)
Now, divide both sides by 146,000:
\[ \gamma - 1 = \frac{-58100}{146000} \]
Calculating the right-hand side:
\[ \gamma - 1 \approx -0.3973 \]

### Step 6: Solve for \(\gamma\)
Now, add 1 to both sides:
\[ \gamma \approx 1 - 0.3973 \approx 0.6027 \]

### Step 7: Interpret the value of \(\gamma\)
Since \(\gamma\) is typically a positive value greater than 1, we need to ensure our calculations align with the physical properties of gases. The value calculated seems incorrect due to the sign. Let's correct it:
\[ \gamma - 1 = \frac{58100}{146000} \implies \gamma \approx 1 + 0.3973 \approx 1.3973 \]

### Step 8: Identify the type of gas
The value of \(\gamma\) is approximately 1.4, which is characteristic of a diatomic gas (like O2 or N2).

### Final Answer
The gas is diatomic.

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