Question

Three moles of a ideal gas at 200K and 2.0 atm pressure undergo reversible adiabatic compression until the temperature becomes 250K for the gas Cvis 27.5JK-1mol-1in this temperature range. Calculate the q , w ,delta U , delta H & final V and P. Given (5/4)1/0.3= 2.1 , (5/4)13/3= 2.61

Answer 1

Explanation:

Solution

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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, CP−CV=R

∴CP=R+CV=8.314+27.5=35.814 J mol−1K−1

∴△H=nCP△T=3×35.814×50=5372.1J