An ideal, monatomic gas is allowed to expand quasi-statically and adiabatically

Question

An ideal, monatomic gas is allowed to expand quasi-statically and adiabatically from an initial volume Vi = 4.1 liter and initial temperature Ti = 390 K to a final volume of Vf = 9.2 liter. If there are n= 0.8 mol of gas, find the final temperature Tf. What is the final pressure pf?What is the work done by the gas?What is the change in the internal energy of the gas?Now, suppose that the gas expands quasi-statically and isothermally. Using the same values for Vi, Vf, Ti, and n, calculate the final pressure.Calculate the heat added to the system in this isothermal expansion.

Explanation / Answer

for quasistatic adabatic process

PV^=const

=5/3 for monoatomic gas

the equation can also be written as

TiVi^(-1)=TfVf^(-1)

390*4.1^2/3=Tf*9.2^2/3

Tf=999.04/4.3906=227.54K

Pf= 0.8*.082*227.54/9.2=1.62atm

as

Q=U+W

U=nCvT

change in internal energy=0.8*(3/2)*8.314*(227.54-390)=-1620.83J(-ve)

workdone=-U=-1620.83J

in isothermal process

T=0

PiVi=PfVf

.8*.082*390=Pf*9.2

Pf=2.7atm

Q=U+w

for isothermal U=0 as T=0

Q=w=.8*8.314*390ln(9.2/4.1)=2096.48J

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