During an adiabatic expansion the temperature of 0.440 mol of argon (ar) drops from 61.0 ∘c to 10.0 ∘c. the argon may be treated as an ideal gas.(a) draw a pv-diagram for this process.
(b) how much work does the gas do?
(c) what is the change in internal energy of the gas? explain.

W= 2.79 atm.L

ΔU= -2.79 atm.L

Explanation:

Given that

T₁ = 61⁰C

T₂=10⁰C

n= 0.44 moles

We know that specific heat capacity ratio for Ar

γ = 1.66

R=gas constant = 0.0821 atm.L/K.mol

The work done given as

W= 2.79 atm.L

Given that process is adiabatic that is why heat transfer is zero.

From first law

Q= W + ΔU

ΔU=Change in the internal energy

Q=Heat transfer

W= Work transfer

0 = 2.79 + ΔU

ΔU= -2.79 atm.L

v=ir = 3x15 =45 v