A 0.460-mol sample of an ideal diatomic gas at 404 kPa and 276 K expands quasi-s
ID: 1480485 • Letter: A
Question
A 0.460-mol sample of an ideal diatomic gas at 404 kPa and 276 K expands quasi-statically until the pressure decreases to 168 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is the following.
(a) isothermal
final temperature__________J
volume of the gas_______L
work done by the gas_________J
heat absorbed___________J
(b) adiabatic
final temperature__________K
volume of the gas___________L
work done by the gas_________J
final temperature__________J
volume of the gas_______L
work done by the gas_________J
heat absorbed___________J
(b) adiabatic
final temperature__________K
volume of the gas___________L
work done by the gas_________J
heat absorbed_________JExplanation / Answer
isothermal process:
temperature remains constant
number of moles=n=0.46 moles
initial pressure=P1=404*10^3 Pa
initial temperature=T1=276 K
final pressure=168*10^3 Pa
initial volume=number of moles*R*initial temperature/initial pressure
=0.46*8.314*276/(404*10^3)=2.6127*10^(-3) m^3
as temperature remains constant, final temperature=initial temperature
=276 K
using ideal gas equation, as temperature constant,
pressure*volume=constant
==>initial pressure*initial volume=final pressure*final volume
==>404*10^3*2.6127*10^(-3)=168*10^3*final volume
==>final volume=404*10^3*2.6127*10^(-3)/(168*10^3)
=6.283*10^(-3) m^3
work done by the gas=-number of moles*R*T*ln(final volume/initial volume)
=-0.46*8.314*276*ln(6.283/2.6127)=-926.2 J
as there is no change in temperature, change in intenral energy is also 0
hence heat absorbed=work done on the gas
=-work done by the gas
=-(-926.2)=926.2 J
part b:
in adiabatic process,
no heat is transferred from the surrounding to the system and vice versa.
part 1:
relationship between temperature and pressure is given by
pressure^(1-gamma)*temperature^(gamma)=constant
where gamma=Cp/Cv=1.4 for a diatomic gas
hence (404*10^3)^(1-1.4)*276^1.4=(168*10^3)^(1-1.4)*final temperature^1.4
==>final temperature^1.4=1840.11
==>final temperature=214.798 K
part 2:
using final temperature and pressure values in P*V=n*R*T ideal gas equation:
168*10^3*final volume=0.46*8.314*214.798
==>final volume=4.8897*10^(-3) m^3
part 3:
work done by the gas=(final pressure*final volume-initial pressure*initial volume)/(1-gamma)
=(168*10^3*4.8897*10^(-3)-404*10^3*2.6127*10^(-3))/(1-1.4)=585.153 J
part 4:
as in adiabatic process there is no heat transfer, heat absorbed=0
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