Question

A cylinder of volume \( 0.1 \, {m}^3 \) is filled with nitrogen at 10 MPa and 300 K. Consider nitrogen to be an ideal gas. The cylinder develops a leak and nitrogen escapes to atmosphere which is at 0.1 MPa. After some time, the pressure in the cylinder reduces to 5 MPa. Assuming the cylinder and the leaked gas temperature remains constant at 300 K, the work done (in MJ) by nitrogen gas is

• A. 0.1
• B. 1
• C. 0.5
• D. 10

Hint:

In isothermal processes, the work done by or on the gas can be calculated using the formula \( W = P_1 V_1 \ln \left( \frac{V_2}{V_1} \right) \) when the temperature remains constant.

Answer 1

The Correct Option is C

Given:
- Volume of the cylinder, \( V = 0.1 \, {m}^3 \)
- Initial pressure, \( P_1 = 10 \, {MPa} \)
- Final pressure, \( P_2 = 5 \, {MPa} \)
- Temperature, \( T = 300 \, {K} \)
- The nitrogen behaves as an ideal gas.
For an ideal gas, the work done during an isothermal process is given by the equation: \[ W = nRT \ln \left( \frac{V_2}{V_1} \right) \] where \( n \) is the number of moles, \( R \) is the gas constant, and \( V_1 \) and \( V_2 \) are the initial and final volumes. From the ideal gas law: \[ P_1 V_1 = nRT {and} P_2 V_2 = nRT \] Since \( T \) and \( n \) are constant, we can write: \[ \frac{V_2}{V_1} = \frac{P_1}{P_2} \] Substituting the known values: \[ \frac{V_2}{V_1} = \frac{10}{5} = 2 \] Now, the work done by the gas is: \[ W = nRT \ln \left( 2 \right) \] Since \( nRT = P_1 V_1 \), we can substitute: \[ W = P_1 V_1 \ln \left( 2 \right) \] Substituting the known values: \[ W = 10 \times 10^6 \times 0.1 \times \ln \left( 2 \right) \approx 0.5 \, {MJ} \] Thus, the work done by the nitrogen gas is \( 0.5 \, {MJ} \).