Two moles of an ideal monoatomic gas at 5 bar and 300 K are expanded irreversibly up to a final pressure of 1 bar and 240 K against an external pressure of 0.5 bar. The work done by the gas is `-xR`. The value of x' is
(Here 'R' is gas constant)

Answer

To solve the problem, we need to calculate the work done by the gas during its irreversible expansion. Let's break down the steps:

### Step 1: Calculate Initial Volume (V1)
Using the ideal gas equation \( PV = nRT \):

1. Given:
- Number of moles \( n = 2 \)
- Initial pressure \( P_1 = 5 \) bar
- Initial temperature \( T_1 = 300 \) K
- Gas constant \( R \) (we will keep it as \( R \))

2. Rearranging the equation:
\[ V_1 = \frac{nRT_1}{P_1} \]

3. Substituting the values:
\[ V_1 = \frac{2 \times R \times 300}{5} \]
\[ V_1 = \frac{600R}{5} = 120R \]

### Step 2: Calculate Final Volume (V2)
Using the same ideal gas equation for the final state:

1. Given:
- Final pressure \( P_2 = 1 \) bar
- Final temperature \( T_2 = 240 \) K

2. Rearranging the equation:
\[ V_2 = \frac{nRT_2}{P_2} \]

3. Substituting the values:
\[ V_2 = \frac{2 \times R \times 240}{1} \]
\[ V_2 = 480R \]

### Step 3: Calculate Change in Volume (ΔV)
Now we find the change in volume:
\[ \Delta V = V_1 - V_2 \]
Substituting the values:
\[ \Delta V = 120R - 480R = -360R \]

### Step 4: Calculate Work Done (W)
The work done by the gas during irreversible expansion against an external pressure is given by:
\[ W = P_{\text{external}} \times \Delta V \]

1. Given:
- External pressure \( P_{\text{external}} = 0.5 \) bar

2. Substituting the values:
\[ W = 0.5 \times (-360R) \]
\[ W = -180R \]

### Step 5: Relate Work Done to the Given Expression
According to the problem, the work done by the gas is given as \( -xR \). We have calculated:
\[ W = -180R \]

Thus, we can equate:
\[ -xR = -180R \]

### Step 6: Solve for x
From the equation above, we find:
\[ x = 180 \]

### Final Answer
The value of \( x \) is \( 180 \).

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