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Certain perfect gas is found to obey PV3/2 = constant during an adiabatic process. If such a gas at initial temperature T is adiabatically compressed to half the initial volume, its final temperature will be
- a)√2 T
- b)2 T
- c)2√2 T
- d)4 T
Correct answer is option 'A'. Can you explain this answer?
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Certain perfect gas is found to obey PV3/2= constantduring an adiabati...
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Certain perfect gas is found to obey PV3/2= constantduring an adiabati...
T/2
b) T
c) 2T
d) 4T
Given that PV^3/2 = constant for the gas during an adiabatic process, we have:
P1V1^3/2 = P2V2^3/2
Since the process is adiabatic, we have:
P1V1^γ = P2V2^γ
where γ is the ratio of specific heats for the gas.
Since the gas is being adiabatically compressed to half the initial volume, we have:
V2 = V1/2
Substitute this into the equation for the adiabatic process:
P1V1^γ = P2(V1/2)^γ
Solving for P2:
P2 = P1(1/2)^(3/γ) = P1(1/2)^(3/5)
Now, for an ideal gas, the pressure and temperature are directly proportional during an adiabatic process:
P1/T1 = P2/T2
Substitute the expressions for P2 and V2:
P1/T1 = P1(1/2)^(3/5) / T2
Solving for T2:
T2 = T1/(1/2)^(3/5) = T1/(2^3/5) = T1/(2^(3/5)) = T1/2^0.6
Therefore, the final temperature after adiabatic compression to half the initial volume will be T/2^(0.6) = T/2^0.6 = T/2^3/5 = T/2^(3/5) = T/2^(0.6) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T
b) T
c) 2T
d) 4T
Given that PV^3/2 = constant for the gas during an adiabatic process, we have:
P1V1^3/2 = P2V2^3/2
Since the process is adiabatic, we have:
P1V1^γ = P2V2^γ
where γ is the ratio of specific heats for the gas.
Since the gas is being adiabatically compressed to half the initial volume, we have:
V2 = V1/2
Substitute this into the equation for the adiabatic process:
P1V1^γ = P2(V1/2)^γ
Solving for P2:
P2 = P1(1/2)^(3/γ) = P1(1/2)^(3/5)
Now, for an ideal gas, the pressure and temperature are directly proportional during an adiabatic process:
P1/T1 = P2/T2
Substitute the expressions for P2 and V2:
P1/T1 = P1(1/2)^(3/5) / T2
Solving for T2:
T2 = T1/(1/2)^(3/5) = T1/(2^3/5) = T1/(2^(3/5)) = T1/2^0.6
Therefore, the final temperature after adiabatic compression to half the initial volume will be T/2^(0.6) = T/2^0.6 = T/2^3/5 = T/2^(3/5) = T/2^(0.6) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T/2^(3/5) = T
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Certain perfect gas is found to obey PV3/2= constantduring an adiabatic process. If such a gas at initialtemperature T is adiabatically compressed to half theinitial volume, its final temperature will bea)√2 Tb)2 Tc)2√2 Td)4 TCorrect answer is option 'A'. Can you explain this answer?
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