Question

One mole of monoatomic gas expands adiabatically at initail temperature t against a constant pressure of 1 atm from 1l to 2l. Find out final temperature

Answer 1

Answer: T - 23×0.0821

Here Pext=1atm

V1 = 1 L and V2 = 2 L

R = 0.0821 L atm K−1mol−1

γ=5/3 (For a monoatomic gas)

T!=T K and T2=?

Now, CP−CV=R or CPCV−1=RCV

⇒γ−1=RCV

CV=Rγ−1

⇒CV(T2−T1)=Pext(V1−V2)

⇒Rγ−1(T2−T1)=Pext(V1−V2)

Substituting the values,

0.08215/3−1(T2−T1)=1×(1−2)

⇒T2−T=2/30.0821

⇒T2=T−23×0.0821

Hope this answer helps you

Please mark it as the brainliest answer

Answer 2

hope u understand .............