Two moles of an ideal monoatomic gas at 5 bar and 300 K are expanded irreversibly up to an external pressure of 1 bar. The work is done by the gas is xR (in bar. Litre). The value of 'x' is [R is gas constant]

Answer

To solve the problem step by step, we will follow the process of calculating the work done by the gas during its irreversible expansion.

### Step 1: Identify the Given Data
- Number of moles of gas (n) = 2 moles
- Initial pressure (P1) = 5 bar
- Final external pressure (P_ext) = 1 bar
- Initial temperature (T1) = 300 K
- Gas constant (R) = 0.0831 L·bar/K·mol (or 8.314 J/K·mol)

### Step 2: Calculate the Final Temperature (T2)
For an ideal gas, we can use the relation for the change in temperature during an irreversible process. The work done (W) can be expressed as:

\[ W = -P_{ext} \cdot (V_2 - V_1) \]

To find the volumes \( V_1 \) and \( V_2 \), we can use the ideal gas law:

1. Calculate \( V_1 \) using \( P_1 \) and \( T_1 \):
\[ V_1 = \frac{nRT_1}{P_1} = \frac{2 \cdot R \cdot 300}{5} \]

2. Calculate \( V_2 \) using \( P_{ext} \) and \( T_2 \):
\[ V_2 = \frac{nRT_2}{P_{ext}} = \frac{2 \cdot R \cdot T_2}{1} \]

### Step 3: Relate T2 to the Work Done
Using the first law of thermodynamics, we can relate the work done to the change in internal energy and heat transfer. For a monoatomic ideal gas, the change in internal energy \( \Delta U \) can be expressed as:

\[ \Delta U = nC_v(T_2 - T_1) \]

Where \( C_v \) for a monoatomic gas is \( \frac{3}{2}R \). Thus:

\[ \Delta U = 2 \cdot \frac{3}{2}R(T_2 - 300) \]

### Step 4: Substitute for Work Done
From the first law of thermodynamics, we know:

\[ W = -P_{ext}(V_2 - V_1) \]

Substituting the expressions for \( V_1 \) and \( V_2 \):

\[ W = -1 \cdot \left( \frac{2RT_2}{1} - \frac{2R \cdot 300}{5} \right) \]

### Step 5: Solve for T2
Now we can set up the equation for work done:

\[ W = -1 \cdot \left( 2RT_2 - \frac{2R \cdot 300}{5} \right) \]

This simplifies to:

\[ W = -2R(T_2 - 60) \]

### Step 6: Calculate Work Done
Now we can equate the expressions for \( W \):

\[ W = -2R(T_2 - 60) \]

To find \( T_2 \), we need to solve this equation. We can rearrange it to find \( T_2 \):

1. Calculate \( W \) using \( P_{ext} \) and the volumes:
\[ W = -1 \cdot \left( \frac{2RT_2}{1} - \frac{120R}{1} \right) \]

2. Set the above equal to the expression for work done in terms of temperature.

### Step 7: Final Calculation
After calculating, we find:

\[ T_2 = 204 K \]

### Step 8: Calculate Work Done
Finally, substitute \( T_2 \) back into the work done formula:

\[ W = -1 \cdot \left( 2R \cdot 204 - 120R \right) = -1 \cdot (408R - 120R) = -288R \]

### Conclusion
The value of \( x \) in the expression \( W = xR \) is:

\[ x = -288 \]