Question

An ideal gas initially at 350 K undergoes an isobaric expansion at 2.50 kPa. The volume increases from 1.00 m3 to 3.00 m3 and 13.0 kJ is transferred to the gas by heat. (a) What is the change in internal energy of the gas? kJ (b) What is the final temperature of the gas?

Answer 1

Change of Internal energy is sum of heat transferred and work done on the gas
U = Q + W

Heat transferred, Q = 13.0KJ
Work done on the gas by changing Volume from V₁ to V₂ for isobaric process ( p=constant)
W = - p*(V₂ - V₁)
for this problem
W = - 2.5×10³Pa · (3.0m³ - 1m³) = -5kJ
Therefore
U = 15kJ - 5kJ = 10kJ

(b)
Use Ideal gas law
p·V = N·R·T
Since pressure p and N stay constant throughout the process:
V/T = (NR)/p= constant
hence:
V₁/T₁ = V₂/T₂

T₂ = (V₂/V₁)T₁
= (3m³/1m³)*350K
= 1050K