# One gram of oxygen (О₂) is heated from Т₁ = 283K to Т₂ = 333K in various ways: a) isobaric;

**One gram of oxygen (О₂) is heated from Т₁ = 283K to Т₂ = 333K in various ways: a) isobaric; b) isochoric. Find the change in ΔU and Q supplied to oxygen when it is heated from Т₁ to Т₂, in each case.**

Given

mass of oxygen (O2) = 1 gram

T1 = 283 K

T2 = 333 K

a) isobaric process V = const, A = 0

Q = c * m * (T2 – T1) = ΔU, where c is the specific heat at a constant volume of 0.157 kcal / (kg * deg)

then Q = 0, 157 * 10 to the power (- 3) * (333 – 283) J = 7, 85 * 10 to the power (- 3) J = 0, 00785 J

2) isochoric process P = const

Q = A + ΔU

Q = Cp * m * ΔT, where Cp is the specific heat at constant pressure = 0.22 kcal / (kg * deg)

that is, Q = 0. 22 * 10 to the power (- 3) * (333 – 283) J = 11 * 10 to the power (- 3) J = 0, 011 J

find A

A = m * R * ΔT / M = 10 to the power (- 3) * 8, 31 * 50/32 * 10 to the power (- 3) J = 8, 31 * 50/32 J = 12.99 J

then ΔU = Q – A = 0, 0 11 J – 12, 99 J = – 12, 979 J