Physical chemistry
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~
1 個解答
- 1 0 年前最佳解答
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