A vertical cylinder contains N molecules of a monatomic ideal gas and is closed

Question

A vertical cylinder contains N molecules of a monatomic ideal gas and is closed off at the top by a piston of mass M and area A. The acceleration due to gravity is g. The heat capacities of the piston and cylinder are negligibly small, and any frictional forces between the piston and the cylinder walls can be neglected. The whole system is thermally insulated. Initially, the piston is clamped in position so that the gas has a volume V0 and a temperature T0. The piston is now released and, after some oscillations, comes to rest in a final equilibrium situation corresponding to a larger volume of the gas. Calculate the final temperature of the gas in terms of N, M, A, g, V0, T0, and kB.

Explanation / Answer

Solution) N = number of molecules

M= mass

Area=A

Acceleration due to gravity =g

Volume =Vo

Temperature =To

Final temperature =Tf

Converting molecules to moles

n=N/(6.02×10^(23))

We have from first law

Change in internal energy =Work done

dU=W

nCvdT=W

But work done W=Mgh

W=(Mg/A)(Ah)

W=(Mg(Vf-Vo)/A)

Work done on the gas so W is negative

nCvdT=(-)(Mg/A)(Vf-Vo)

nCv(Tf-To)=(MgVo/A)-(MgVf/A)

For Vf we have final pressure at equilibrium

Pf=(Mg/A)

(Mg/A)=(nRTf/Vf)

Vf=nARTf/(Mg)

nCv(Tf-To)=(Mg/A)-(Mg/A)(nRTfA/Mg)

nCvTf-nCvTo=(MgVo/A)-nRTf

(nCv+nR)Tf- nCvTo=(MgVo/A)

Tf=(CvTo+(MgVo/nA))(1/(Cv+R))

Tf=(CvTo+(MgVo(6.02×10^(23)/NA)(1/(Cv+NKB))

From nR=NKB

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