317 mol of gas undergoes the process 1 2 shown in the figure

3.17 mol of gas undergoes the process 1--> 2 shown in the figure. What is the temperature at point 2? Please provide final answer, thanks

p (atm) . 0 V (cm3) 0 1000 2000 3000

Solution

Initial point 1

P = 3 atm

V = 1000 cm³

PV = 3000

Final point 2

P = 1 atm

V = 3000 cm³

PV = 3000

Hence we see that PV = constant .

I.e the process is isothermal .

Hence temperetue will be constant.

PV = nRT

(1.01325×10⁵)×(3×10^-3)= 3.17×8.3145×T

Or. T = 11.53 K = -261.47°C