Question
Suppose 3 mols of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume. Find the highest temperature attained by the gas. Find the lowest temperature attained by the gas. Find the highest pressure attained by the gas. Find the lowest pressure attained by the gas.
Answers
In an isothermal process, P*V = contant
In an adiabatic process (at constant entropy), P*V^gamma = constant
where gamma is the specific heat ratio, Cp/Cv. (Adiabatic expansions are isentropic unless they are into a vacuum and do no work)
For neon and other monatomic gases,
gamma = 5/3.
Original To = 273 K
Original Po = 1.00 atm
Original Vo = 3*22.4 = 67.2 liters
After the isothermal compression,
V1 = 12.77 liter = 0.19 Vo
T1 = 273 K
P1 = (1/0.19)Po = 5.26 Po
After the adiabatic expansion,
V2 = Vo
P2*V2^5/3 = P1*V1^5/3
P2/P1 = (V1/V2)^5/3 = (V1/V0)^5/3
= 0.19^5/3 = 0.0628
P2 = 0.0628 *5.26 = 0.330 atm
P2*V2/T2 = Po*Vo/To
T2 = (P2/Po)(V2/Vo)*To
= 0.330* 273 = 90.2 K
Highest P = Po = 5.26 atm
Lowest P = P2 = 0.330 atm
Highest T = To
Lowest T = T2
In an adiabatic process (at constant entropy), P*V^gamma = constant
where gamma is the specific heat ratio, Cp/Cv. (Adiabatic expansions are isentropic unless they are into a vacuum and do no work)
For neon and other monatomic gases,
gamma = 5/3.
Original To = 273 K
Original Po = 1.00 atm
Original Vo = 3*22.4 = 67.2 liters
After the isothermal compression,
V1 = 12.77 liter = 0.19 Vo
T1 = 273 K
P1 = (1/0.19)Po = 5.26 Po
After the adiabatic expansion,
V2 = Vo
P2*V2^5/3 = P1*V1^5/3
P2/P1 = (V1/V2)^5/3 = (V1/V0)^5/3
= 0.19^5/3 = 0.0628
P2 = 0.0628 *5.26 = 0.330 atm
P2*V2/T2 = Po*Vo/To
T2 = (P2/Po)(V2/Vo)*To
= 0.330* 273 = 90.2 K
Highest P = Po = 5.26 atm
Lowest P = P2 = 0.330 atm
Highest T = To
Lowest T = T2
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