Question

One mole of an ideal gas with is compressed adiabatically in a single stage with a constant opposing pressure equal to 10atm. pressure is 10 atm. Calculate the final temperature of the gas, w, q, ΔU and ΔH. HINT – this is not reversible expansion.

Answer 1

a) Heat, Q

Since its an adiabatic process, there is no trnafer of heat, therefore q = 0.

b) Final temperature

Assumption monoatomic gas

= 5/3

Intial T = 273 K, Initiial P = 1 bar (Since 1 mole of ideal at standard temperature & pressure)

Final P = 10 atm = 10.1325 bar, Final Temp = T

For adiabatic process

P^-2/3T^5/3 = constant

1^-2/3(273)5/3= (10.1325)^-2/3T^5/3

T = 689.36 K

c) Change in Internal Energy

= degrees of freedom/ 2 = 3/2 = 1.5

n = 1

= 689.36 - 273 = 416.36 K

= 1.5*8.314*416.36 = 5192.42 J

d) Work

Here Q = 0

= - 5192.42 J