Question

A gas is compressed at a constant pressure of 0.800 atm from 9.00 L to 2.00 L. In the process, 350 J of energy leaves the gas by heat.

(a) What is the work done on the gas?
J

(b) What is the change in its internal energy?
J

Answer 1

Given that :

initial volume, V₀ = 9 L

final volume, V_f = 2 L

heat, Q = -350 J

constant pressure, P = 0.8 atm

(a) The work done on the gas which will be given as :

using an equation,   W = -P V                                     { eq.1 }

where, V = change in volume = V_f - V₀

inserting the values in above eq.

W = - [(0.8 atm) (1.013 x 10⁵ Pa/atm)] [(2 L) - (9 L)] (1 x 10^-3 m³/L)

W = -(0.8104 x 10⁵ Pa) (-7 x 10^-3 m³)

W = 5.6728 x 10² J

W = 567.2 J

(b) The change in its internal energy which willbe given as :

using an equation,   U = Q + W                                         { eq.2 }

inserting the values in eq.2,

U = (-350 J) + (567.2 J)

U = 217.2 J