Question

Two different gases occupy two separate bulbs. (Figure 1) Consider the process that occurs when the stopcock separating the gases is opened, assuming the gases behave ideally.

Part A

Predict the sign of ΔH for the process.

Predict the sign of  for the process.

	ΔH > 0
	ΔH < 0
	ΔH = 0

Part B

Predict the sign of ΔS for the process.

Predict the sign of  for the process.

	ΔS > 0
	ΔS < 0
	ΔS = 0

Part C

Is the process that occurs when the stopcock is opened a reversible one?

Is the process that occurs when the stopcock is opened a reversible one?

	yes
	no

Part D

How does the process affect the enthalpy of the surroundings?

How does the process affect the enthalpy of the surroundings?

	ΔHsurr > 0
	ΔHsurr < 0
	ΔHsurr = 0

Answer 1

ΔH_mix = 0. Because the enthalpy of mixing for two ideal gasses is 0.

ΔS_mix =-nR (x₁ ln x₁ + x₂ ln x₂). In this expression x₁ and x₂ are positive numbers lying between 0 &1. Thus ln x₁ and ln x₂ must be negative numbers. Therefore entropy change in this process is positive. So, ΔS > 0.

The Gibb's free energy change of mixing is calculated as ΔG_mix = ΔH_mix -TΔS_mix.

Now, ΔH_mix = 0 and ΔS_mix > 0. Thus, ΔG_mix < 0 for all temperature. As the process is spontaneous the mixing can not be reversible.

As the enthalpy change of the system is 0 that of the surrounding is also 0. So, ΔH_surr = 0