An ideal gas initially at 335 K undergoes an isobaric expansion at 2.50 kPa. ?
An ideal gas initially at 335 K undergoes an isobaric expansion at 2.50 kPa. The volume increases from 1.00 m3 to 3.00 m3 and 10.6 kJ is transferred to the gas by heat.
(a) What is the change in internal energy of the gas?
(b) What is the final temperature of the gas?
- jacob sLv 72 months ago
An ideal gas initially at 300 K undergoes an isobaric expansion at 2.50 kPa.
(a) If the volume increases from 1.00 m3 to 3.00 m3 and if 15.0 kJ is transferred to the gas by heat, what is the change in its internal energy?
(b) What is its final temperature?
Change of Internal energy equals heat transferred to plus work done on the gas
ΔU = Q + W
Work done on the gas by changing Volume from V₁ to V₂ is given by the integral
W = - ∫ p dV from V₁ to V₂
for isobaric process ( p=constant)
W = - p·(V₂ - V₁)
for this problem
W = - 2.5×10^3Pa · (3m³ - 1m³) = -5kJ
ΔU = 10.6kJ - 5kJ = 5.6kJ
Use ideal gas law
p·V = N·R·T
Since pressure p and N stay constant throughout the process:
V·T = p / (N·R) = constant
V₁·T₁ = V₂·T₂
T₂ = T₁· (V₁/V₂)
= 335K · (1m³ / 3m³)
= 111.67KSource(s): schmiso