Question

3 moles of an ideal gas at 200 K is
compressed reversibly and adiabatically
until final temperature reaches 250 K. If
molar heat capacity at constant volume is
27 cal/mol.K. Find work.​

Answer 1

Explanation:

200/3

=66.66

250/27

=9.2°

Answer 2

Answer:

q=0 (Adiabatic process)

n=3 moles

∴△U=nCV △T=2×27.5×50

∴△U=4125J

Now, △U=q+w

∴△U=w

∴w=4125J

Now, C P−C V =R

∴CP =R+C V =8.314+27.5=35.814 J mol −1 K ∴△H=nCP△T=3×35.814×50=5372.1J