You have two containers of a monatomic ideal gas that are intially at the same P

Question

You have two containers of a monatomic ideal gas that are intially at the same Pressure, Volume, and Temperature.

You add 940 J of heat to container 1 in a constant temperature process.

You add 940 J of heat to container 2 in a constant pressure process.

What is the difference in work done in these processes, 1 - 2? In other words, if W1 is the work done in process 1 and W2 is the work done in process 2, what is W1 - W2?

Give you answer in Joules to three significant digits. Do not include units with your answer.

Explanation / Answer

at constant temperature :-

change in internal energy = 0.

hence... Q = W

W1= 940 J

at constant pressure process:-

W2 = Q - (delta U)

= Cp(delta T) - Cv(delta T) = delta T *(Cp-Cv)= R *(delta T)

now Q = Cp (delta T)

(delta T) = Q/Cp = 940/ (5R/2)

hence W2 = R *(2 *940/5R) = 1880/5 = 376 J.

hence W1 - W2 = 940 - 376= 564 J.

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