One mole of an ideal gas with Cv,m=3/2 R initially at 298k and 1.00X10^5 Pa undergoes a reversible adiabtic

Favorites|Homepage
Subscriptions | sitemap

HOME > > One mole of an ideal gas with Cv,m=3/2 R initially at 298k and 1.00X10^5 Pa undergoes a reversible adiabtic

One mole of an ideal gas with Cv,m=3/2 R initially at 298k and 1.00X10^5 Pa undergoes a reversible adiabtic

[From: ] [author: ] [Date: 13-09-25] [Hit: ]

w, delta U and delta H for this process.-Uh, this can easily be solved with a closed system energy balance.dKE+dPE+dU = Q - W. For a reversible adiabatic process,......

compression. At the end of the process the pressure is 1.00 x 10^6 Pa. Calculate the final temperature of the gas, q, w, delta U and delta H for this process.

-

Uh, this can easily be solved with a closed system energy balance.

dKE+dPE+dU = Q - W. For a reversible adiabatic process, q = 0. CvdT = dU, so integrate. dH = dU +PdV. PE = 0 since there is no reference frame change. dKE is also zero. So, there you have it!

1

keywords: with,and,ideal,undergoes,gas,an,of,1.00,Pa,mole,10,initially,Cv,adiabtic,at,One,298,reversible,One mole of an ideal gas with Cv,m=3/2 R initially at 298k and 1.00X10^5 Pa undergoes a reversible adiabtic

New

Hot