Question

A monatomic ideal gas that is initially at a pressure of 1.50 ✖️ 10 to the power five as a volume of 0.08 m³ compressed adiabatically to a vol of 0.04 m³

What is the final temp?

How much work is done by the gas?

What is the ratio of final temp of the gas to its final temp?

Answer 1

Explanation:

pressure was released suddenly, so that your 3rd suggestion is correct. With an irreversible expansion against a constant external pressure, the adiabatic equation pVγ=const is inapplicable, so one must use the first law of thermodynamics. Work is done at the expense of the loss of internal energy:

W=−ΔU=nCv(T1−T2)

For ideal monoatomic gas the following is applied:

n=

p1V1

RT1

Cv=

3R

2

so that (1) can be rewritten as such:

W=nCv(T1−T2)=

3p1V1R

2RT1

(T1−T2)=

3p1V1

2

−

3p1V1T2

2T1

At the same time work by gas expansion is

W=p2(V2−V1)=p2V2−p2V1

and from the ideal gas law the reduced product is

p2V2=nRT2

so that (2) can also be rewritten:

W=p2(V2−V1)=nRT2−p2V1

I would leave for OP the task to to equate (1.3) and (2.2) and express the final temperature T2 from there:

3p1V1

2

−

3p1V1T2

2T1

=nRT2−p2V1

Here I bring up the final result:

T2=

3p1V1+2p2V1

5p1V1

T1=

3⋅10 atm⋅10 L+2⋅1 atm⋅10 L

5⋅10 atm⋅10 L

⋅273.15 K=174.7 K

Volume after expansion is also to be found using ideal gas law and (1.1):

V2=

nRT2

p2

=

p1V1RT2

RT1p2

=

p1T2

p2T1

V1=

10 atm⋅174.7 K

1 atm

Answer 2

Heya Mate........

Your answer in attachment.....

Hope this will help you.....