An ideal gas expands at a constant total pressure of 3.3 atm from 500 mL to 830

Question

An ideal gas expands at a constant total pressure of 3.3 atm from 500 mL to 830 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.

Part A: 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 = - 3.3*101325Pa (830×10m³ - 500×10m³) = -110.34J

That means the gas does 110.34J 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 = - (-110.34j) = 110.34J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.