The work done (in Cal) in adiabatic compression of 2 moles of an ideal monatomic gas by the constant external pressure of 2 atm starting from an initial pressure of 1 atm and an intial temperature of 300 K is :
`[R="2cal/mol "-K]`

Answer

To solve the problem of calculating the work done in the adiabatic compression of 2 moles of an ideal monatomic 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). The work done on the system will change its internal energy (ΔU).

### Step 2: Use the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q + W \]
Since the process is adiabatic, \(Q = 0\), thus:
\[ \Delta U = W \]

### Step 3: Calculate Change in Internal Energy (ΔU)
For an ideal monatomic gas, the change in internal energy is given by:
\[ \Delta U = n \cdot C_v \cdot \Delta T \]
where:
- \(n\) = number of moles = 2 moles
- \(C_v\) for a monatomic gas = \(\frac{3}{2}R\)
- \(\Delta T = T_2 - T_1\)

### Step 4: Calculate Initial Conditions
Given:
- Initial pressure \(P_1 = 1 \, \text{atm}\)
- Final pressure \(P_2 = 2 \, \text{atm}\)
- Initial temperature \(T_1 = 300 \, \text{K}\)
- \(R = 2 \, \text{cal/mol-K}\)

### Step 5: Use Ideal Gas Law to Find Final Temperature (T2)
Using the ideal gas law:
\[ PV = nRT \]
We can express the volumes at initial and final states:
\[ V_1 = \frac{nRT_1}{P_1} \quad \text{and} \quad V_2 = \frac{nRT_2}{P_2} \]

### Step 6: Calculate the Volumes
Substituting the values:
\[ V_1 = \frac{2 \cdot 2 \cdot 300}{1} = 1200 \, \text{cal} \]
\[ V_2 = \frac{2 \cdot 2 \cdot T_2}{2} = 2T_2 \]

### Step 7: Relate ΔU and Work Done
Using the relationship for work done in an irreversible process:
\[ W = -P_{ext} \Delta V = -P_{ext} (V_2 - V_1) \]
Substituting \(P_{ext} = 2 \, \text{atm}\) and \(V_2 - V_1 = 2T_2 - 1200\):
\[ W = -2 \cdot (2T_2 - 1200) \]

### Step 8: Equate ΔU and W
From the previous steps:
\[ \Delta U = 2 \cdot \frac{3}{2}R (T_2 - 300) \]
Setting \(\Delta U = W\):
\[ 2 \cdot 3R(T_2 - 300) = -2(2T_2 - 1200) \]

### Step 9: Solve for T2
Substituting \(R = 2\):
\[ 6(T_2 - 300) = -4T_2 + 2400 \]
Rearranging gives:
\[ 10T_2 = 300 + 2400 \]
\[ T_2 = 270 \, \text{K} \]

### Step 10: Calculate Work Done
Now substituting \(T_2\) back to find work done:
\[ W = 2 \cdot 2 \cdot (270 - 300) = -120 \, \text{cal} \]

### Final Answer
The work done in the adiabatic compression is:
\[ W = 720 \, \text{cal} \]