Asked by nirali
An ideal monatomic gas initially has a temperature of 333 K and a pressure of 7.15 atm. It is to expand from volume 581 cm3 to volume 1360 cm3. If the expansion is isothermal, what are (a) the final pressure (in atm) and (b) the work done by the gas? If, instead, the expansion is adiabatic, what are (c) the final pressure (in atm) and (d) the work done by the gas?
Answers
Answered by
drwls
In the isothermal case, P*V = constant = P1 V1.
The work done is Integral P dV
= Integral (P1 V1)dV/V
= P1 V1 ln(V2/V1)
In the adiabatic case,
P*V^(5/3) = constant
(for monatomic gases)
The work is once again Integral P dV, but you get a different answer.
Work = [P1*V1^(5/3)]*Integral dV/V^(5/3)
The work done is Integral P dV
= Integral (P1 V1)dV/V
= P1 V1 ln(V2/V1)
In the adiabatic case,
P*V^(5/3) = constant
(for monatomic gases)
The work is once again Integral P dV, but you get a different answer.
Work = [P1*V1^(5/3)]*Integral dV/V^(5/3)
Answered by
nirali
hey drwl thank u so much. but i didn't get right answer for work done. do i hav to convert pressure to pascal for work done? cn u help me?
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