Two litres of an ideal gas at a pressure of 10 atm expands isothermally into vacuum until its total volume is 10 litres.

• 1. How much heat is absorbed and how much work is done in the expansion.
• 2. If the same expansion takes place against a constant external pressure of 1 atm?
• 3. If the same expansion takes place to a final volume of 10 litres conducted reversibly?

To analyze the expansion of an ideal gas under different conditions, we can break down the problem into three parts.

1. Isothermal Expansion into Vacuum

When an ideal gas expands into a vacuum, no work is done because there is no opposing pressure. The heat absorbed during isothermal expansion can be calculated using the formula:

• Q = nRT ln(Vf/Vi)

Here, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, Vi is the initial volume (2 L), and Vf is the final volume (10 L). Since the process is isothermal, the temperature remains constant. However, in a vacuum, the work done (W) is:

• W = 0

Thus, the heat absorbed is equal to the internal energy change, which is also zero for an ideal gas in this scenario.

2. Expansion Against a Constant External Pressure

When the gas expands against a constant external pressure of 1 atm, work is done by the gas. The work done can be calculated using:

• W = -Pext (Vf - Vi)

Substituting the values:

• W = -1 atm × (10 L - 2 L) = -8 atm·L

To convert this to Joules (1 atm·L = 101.325 J), we get:

• W = -8 × 101.325 J = -810.6 J

The heat absorbed can be calculated using the first law of thermodynamics:

• Q = ΔU + W

Since the internal energy change (ΔU) for an ideal gas in this case is zero (no temperature change), we find:

• Q = 0 - 810.6 J = -810.6 J

3. Reversible Expansion to 10 Litres

For a reversible expansion, the work done can be calculated using:

• W = -nRT ln(Vf/Vi)

Assuming the temperature is constant, we can calculate the work done. The heat absorbed will be equal to the work done in this case:

• Q = -W

Using the same formula as before, we can find:

• W = -nRT ln(10/2)

Thus, the heat absorbed will also be:

• Q = nRT ln(10/2)

In summary, the results for each scenario are:

• Isothermal expansion into vacuum: Q = 0, W = 0
• Expansion against 1 atm: Q = -810.6 J, W = -810.6 J
• Reversible expansion: Q = nRT ln(5), W = -nRT ln(5)