An ideal gas resides in a closed cylinder (diameter is 0.5 ft) with a frictionless piston. The initial conditions are 139 mol of the ideal gas at 25°C. The piston is compressed isothermally to one third of the initial volume. The heat capacity C_v of an ideal gas is 1.5 middot R.

Required:
a. What is the heat interaction (kJ) for this process?
b. The piston now expands isothermally to 120% of the same initial volume. Will the heat interaction increase, decrease, or stay the same?

Question

31-12-2020
Chemistry

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An ideal gas resides in a closed cylinder (diameter is 0.5 ft) with a frictionless piston. The initial conditions are 139 mol of the ideal gas at 25°C. The piston is compressed isothermally to one third of the initial volume. The heat capacity C_v of an ideal gas is 1.5 middot R.

Required:
a. What is the heat interaction (kJ) for this process?
b. The piston now expands isothermally to 120% of the same initial volume. Will the heat interaction increase, decrease, or stay the same?

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sebassandin sebassandin · Answer 1 · 2021-01-06T02:27:02+01:00

Answer:

a. 3.79e5 J

b. Decrease.

Explanation:

Hello!

a. In this case, since the heat involved during a compression-expansion isothermal process is computed via:

[tex]Q=nRTln(\frac{V_2}{V_1} )[/tex]

Now, since the final volume is one third of the initial one:

[tex]V_2=\frac{V_1}{3}[/tex]

So we can plug in now:

[tex]Q=139mol*8.3145\frac{J}{mol*K}*298.15K*ln(\frac{\frac{V_1}{3} }{V_1} )\\\\Q=-3.79x10^5J[/tex]

b. In this case, the relationship between initial and final volume is:

[tex]V_2=2.2V_1[/tex]

So the heat interaction is now:

[tex]Q=139mol*8.3145\frac{J}{mol*K}*298.15K*ln(\frac{2.2V_1 }{V_1} )\\\\Q=2.72x10^5J[/tex]

It means that the heat interaction decrease on the contrary process, it means that in a. heat was released by 3.79e5 J and in b heat is absorbed by 2.72e5 J.

Best regards!