Answer to Question #194108 in Molecular Physics | Thermodynamics for Sostern Munkondya

Question

a certain gas is compressed reversibly from 1 bar , 17 degrees to a pressure of 5bar in a perfectly thermally insulated cylinder the final temperature being 77degrees . The work done on the gas during the compression is 45kj/Kg. Calculate Cv alpha and R

Answer 1

For reversible process$(\frac{P_1}{P_2})^n=\frac{T_1}{T_2}$

$n ln(\frac{P_1}{P_2})=\frac{T_1}{T_2}$

$n=(\gamma-1)/(\gamma)$

$W=R(T_1-T_2)/(\gamma-1)$

Put the Value of data given

$n ln(1/5)=(290/350)$

$n=0.1125$

Again,

$n=(\gamma-1)/\gamma$

$0.1125=(\gamma-1)/(\gamma)$

$\gamma=1.132$

And by work formula

$W=R(T_1-T_2)/(\gamma-1)$

$-45=R(290-350)/(1.1132-1)$

R=0.099KJ/Kg-K

Also $C_v=\frac{R}{(\gamma-1)}$

$C_v=\frac{R}{(1.132-1)}$

Cv=0.75KJ/Kg-K