An ideal gas expands at a constant total pressure of 3.5 atm from 450 mL to 790

Question

An ideal gas expands at a constant total pressure of 3.5 atm from 450 mL to 790 mL . Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value.

Calculate the total work done by the gas in the process.

Express your answer to two significant figures and include the appropriate units.

Part B

Calculate the total heat flow into the gas.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

(a)
In general the work done on the gas is given by the integral
W= - pdV from initial to final volume of the process.
That means no volume change no work is done.
Therefore: only in the first part of the process, in which the gas expands, there is work done.
Since this part is a constant pressure process, the work integral simplifies to:
W = - p dV = - pV

Hence;
W = - 3.5101325Pa (790×10m³ - 450×10m³) = -120.58J

That means the gas does 120.58J of work.


(b)
The internal energy of an ideal gas is function of the temperature:
U = nCvT
Since the gas return to its initial temperature, the total change of internal energy on this whole process is zero:
U = nCvT = 0

On the other hand the change of internal energy equal the sum of heat transferred to the gas plus the work done on it.
U = Q + W = 0

Hence;
Q = - W = - (-120.58) = 120.58J

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