In class, we discussed the case where a gas is initially confined at equilibrium

Question

In class, we discussed the case where a gas is initially confined at equilibrium in a volume V1 at temperature T. A valve is then opened to allow the gas to expand irreversibly and adiabatically (i.e., no heat transfer between system and surroundings) into a second chamber that is initially under vacuum. For the sake of an example, we will consider the specific case where the final volume of the gas is 1000 times larger, so V2 = 1000 V1. Thus, the final state is a system at equilibrium with V2 = 1000 V1 and temperature T. For an ideal gas, this expansion occurs with no change in temperature.

The free expansion case described above is a spontaneous, irreversible process in an isolated system. Thus, the 2nd Law of Thermodynamics assures us that S > 0, but does not tell us what the actual value of the entropy change is.

For 10-3 mole of an ideal gas (n = 0.001 mole), calculate the actual change in entropy (in J/K) involved in moving from state 1 to state 2 for the case where T = 300 K. First, describe a method to provide a reversible path that could carry the system from state 1 to state 2, and then use the approach of question 1 to calculate the change in entropy between state 1 and state 2.

(Note: You should find an answer larger than zero, so the 2nd Law is in good standing as far as this process goes...)

Explanation / Answer

To bring the system from state 1 to 2 reversibly you have to do isothermal reversible expansion. Where a piston is kept at the opening with the same pressure as is on the inside of the container. The pressure of the piston is decreased infinitesmally slowly in finite amount of steps. This way the gas remains in equilibrium throughout the change and it is a reversible change.

The formula for entropy for free expansion at constant temperature is

Delta S= nR ln V/V0

n- moles of gas

R - Ideal gas constant (8.314J/K mol)

V- final volume

V0- Intitial volume

Delta S = 0.001 *8.314* ln (1000) final vol is 1000times initial

= 5.74 × 10-2 J/K

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