A sample of ideal gas (γ = 1.4) is heated at constant pressure. If 140 J of heat is supplied to gas, find ΔU and w.
60 J, -80 J
Do you remember γ value for a diatomic gas?
γ = 1.4
⇒ CPCV = 1.4 ⇒ CP = 1.4 CV
We know that CP − CV = R; 1.4 CV − CV = R
0.4 CV = R
∴ CV = 52R & CP− CV = R; CP − 52R = R
∴ CP = 72R
Now, ΔH = QP = n CP ΔT
ΔT = ΔnCp = 140 × 27n × R = 40n
w = −n R Δ T = −n × 2 × 40n = −80 J
∴ w = −80J
Also
QP = ΔH = ΔU + (−w)
ΔU = ΔH + w
ΔU = 140 − 80
∴ ΔU = 60 J
Also
QP = ΔH = ΔU + (−w)
ΔU = ΔH + w
ΔU = 140 − 80
∴ΔU = 60 J