Question

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?

Answer 1

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:

$Q=nRTln(\frac{V_2}{V_1} )$

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

$V_2=\frac{V_1}{3}$

So we can plug in now:

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

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

$V_2=2.2V_1$

So the heat interaction is now:

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

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!