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 of the work done by the gas during its irreversible expansion, we can follow these steps:

### Step 1: Identify Given Data
- Number of moles (n) = 2 moles
- Initial pressure (P1) = 5 bar
- Initial temperature (T1) = 300 K
- Final pressure (P2) = 1 bar
- Final temperature (T2) = 240 K
- External pressure (P_ext) = 0.5 bar

### Step 2: Calculate Initial Volume (V1)
Using the ideal gas equation \( PV = nRT \):
\[ V_1 = \frac{nRT_1}{P_1} \]
Substituting the values:
\[ V_1 = \frac{2 \text{ moles} \times R \times 300 \text{ K}}{5 \text{ bar}} = \frac{600R}{5} = 120R \]

### Step 3: Calculate Final Volume (V2)
Using the ideal gas equation again for the final state:
\[ V_2 = \frac{nRT_2}{P_2} \]
Substituting the values:
\[ V_2 = \frac{2 \text{ moles} \times R \times 240 \text{ K}}{1 \text{ bar}} = \frac{480R}{1} = 480R \]

### Step 4: Calculate Change in Volume (ΔV)
\[ \Delta V = V_1 - V_2 = 120R - 480R = -360R \]

### Step 5: Calculate Work Done (W)
The work done by the gas during expansion against the external pressure is given by:
\[ W = -P_{\text{ext}} \cdot \Delta V \]
Substituting the values:
\[ W = -0.5 \text{ bar} \cdot (-360R) = 0.5 \cdot 360R = 180R \]

### Step 6: Express Work Done in Terms of R
The work done is given as \( -xR \). From our calculation:
\[ W = 180R \]
Thus, we have:
\[ -xR = 180R \implies x = -180 \]

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

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