Two different gases occupy two separate bulbs. (Figure 1) Consider the process t

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 H for the process.

Predict the sign of  for the process.

Part B

Predict the sign of S for the process.

Predict the sign of  for the process.

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?

Part D

How does the process affect the enthalpy of the surroundings?

How does the process affect the enthalpy of the surroundings?

Explanation / Answer

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

Smix =-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, S > 0.

The Gibb's free energy change of mixing is calculated as Gmix = Hmix -TSmix.

Now, Hmix = 0 and Smix > 0. Thus, Gmix < 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, Hsurr = 0

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