The work done in adiabatic compression of `2` mole of an ideal monoatomic gas by constant external pressure of `2 atm` starting from initial pressure of `1 atm` and initial temperature of `30 K(R=2 cal//"mol-degree")`

Options

A. 36 cal
B. 72 cal
C. 80 cal
D. 100 cal

Explanation

To solve the problem of calculating the work done in the adiabatic compression of 2 moles of an ideal monoatomic gas, we will follow these steps:

### Step 1: Understand the Process
In an adiabatic process, there is no heat exchange with the surroundings (Q = 0). According to the first law of thermodynamics, the change in internal energy (ΔU) is equal to the work done (W) on the system:
\[ \Delta U = W \]

### Step 2: Calculate Change in Internal Energy
For an ideal monoatomic gas, the change in internal energy can be expressed as:
\[ \Delta U = n C_v \Delta T \]
where:
- \( n \) = number of moles (2 moles)
- \( C_v \) = molar heat capacity at constant volume for a monoatomic gas = \( \frac{3}{2} R \)
- \( R \) = gas constant = 2 cal/(mol·K)
- \( \Delta T = T_2 - T_1 \)

### Step 3: Determine Initial and Final Temperatures
We are given:
- Initial temperature \( T_1 = 30 \, K \)
- We need to find the final temperature \( T_2 \).

### Step 4: Use Ideal Gas Law to Relate Pressures and Volumes
Using the ideal gas law, we can express the volumes:
\[ V = \frac{nRT}{P} \]
Thus, we can write:
\[ V_1 = \frac{nRT_1}{P_1} \]
\[ V_2 = \frac{nRT_2}{P_2} \]
where:
- \( P_1 = 1 \, atm \)
- \( P_2 = 2 \, atm \)

### Step 5: Calculate Change in Volume
The change in volume \( \Delta V \) is:
\[ \Delta V = V_2 - V_1 = \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \]

### Step 6: Substitute Values into the Work Equation
The work done on the gas during adiabatic compression against a constant external pressure is given by:
\[ W = -P_{external} \Delta V \]
Substituting \( P_{external} = P_2 \):
\[ W = -P_2 \left( \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \right) \]

### Step 7: Solve for Final Temperature
By equating the expressions for change in internal energy and work done:
\[ n C_v (T_2 - T_1) = -P_2 \left( \frac{nRT_2}{P_2} - \frac{nRT_1}{P_1} \right) \]
Substituting \( C_v \) and simplifying will allow us to solve for \( T_2 \).

### Step 8: Calculate Work Done
Once \( T_2 \) is found, substitute back into the equation for work done:
\[ W = n C_v (T_2 - T_1) \]

### Step 9: Final Calculation
Substituting all known values will yield the final work done in calories.

### Final Answer
After performing the calculations, the work done in the adiabatic compression is found to be \( 72 \, \text{cal} \).

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