Physical chemistry

Question

Initially 0.1 mol of methane is at 1bar pressure and 80℃. The gas behaves ideally and the value of Cp / Cv is 1.31. The gas is allowed to expand reversible and adiabatically to a pressure of 0.1bar.

a.What are the initial and final volumes of the gas?

b.What is the final temperature?

c.Calculate ΔU and ΔH for the process.

請詳解，越詳細越好！感恩～

Answer 1

最佳解答

a.

∵ The gas behaves ideally

∴PV=nRT

initial:

(1bar)Vi=(0.1mol)(0.083)(80+273)

Vi=2.93 (L)

final:

∵The gas behaves ideally,expand reversible and adiabatically

∴PV^(Cp/Cv)=const.

PiVi=PfVf

1*2.93^1.31=0.1*Vf^1.31

Vf=16.99 (L)

b.

∵ The gas behaves ideally

∴PfVf=nRTf

(0.1)(16.99)=(0.1)(0.083)Tf

Tf=204.70 (K)

c.

by lst law

ΔU=Q-W

∵adiabatically

∴Q=0

W=∫PdV

∵The gas behaves ideally,expand reversible and adiabatically

∴PV^(Cp/Cv)=const.

P=const.(V)^(-Cp/Cv)代入積分式

W=∫(const.)V^(-1.31)dV=(const.)/(-0.31)[(Vf)^(-0.31)-Vi^(-0.31)

=(1*10^5pa)(2.93*10^(-3) m^3)^(1.31)/(-0.31)[((2.93*10^(-3))^(-0.31)-(16.99*10^(-3))^(-0.31)]=-397.04 (J)

∴ΔU=Q-W=0-(-3.981)=397.04(J)

ΔH=∫nCpdT

or =ΔU+PΔV+VΔP

(缺Cp數據)

2007-03-30 14:11:39 補充：

a.上面有打錯

∴PV^(Cp/Cv)=const.

PiVi^1.31=PfVf^1.31

參考資料： 自己

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