Question

An ideal-gas process is described by p=cV1/2, where c is a constant.

Part B

0.033 mol of gas at initial temperature of 150∘C is compressed, using this process,from 300 cm3 to 200 cm3. How much work is done on the gas?

Part C

What is the final temperature of the gas in ∘C?

Answer 1

Work done is W = c*(V2^(1-r) - V1^(1-r))/(1-r)

Using P = c*V^(1/2)

P/V^(1/2) = c

P*V^(-1/2) = c

gamma= r = -1/2

1-r =1+1/2 = 3/2

P*V = n*R*T

c*V^(1/2)*V= n*R*T

c*V^(3/2) = n*R*T

c*(300*10^-6)^(3/2) = 0.033*8.314*(150+273)

c = 2.23*10^7

W = c*(V2^(1-r) - V1^(1-r))/(1-r)

W= (2.23*10^7)*((200*10^-6)^(3/2) - (300*10^-6)^(3/2)) / (3/2)

W = -35.2 J

b) using TV^(r-1) = constant

T1*V1^(-3/2) = T2*V2^(-3/2)

(150+273)*300^(-3/2) = T2*200^(-3/2)

T2 = 230.25 K

T2 = 230.25-273 = -42.75 C