An ideal gas expands at a constant total pressure of 3.2 atm from 490 mL to 810

Question

An ideal gas expands at a constant total pressure of 3.2 atm from 490 mL to 810 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) we know that the work done at constant volume
W = P(V2 - V1)
Here V1 = 490 mL = 490*10-6 m3
V2 = 810 *10-6 m3
P = 3.2 atm = 3.2*1.01325*105 N/m2
W = 3.2*1.01325*105(810 - 490)*10-6 = 103.7568 J
(b) We know from first law of thermodynamics
Q = dU+ W
since the temperature drop to its inital value therefore dU = 0
Q = W = 103.7568 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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