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 heat absorbed_________J

Answer 1

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