One mole of a certain ideal gas obtains an amount of heat Q = 1 . 60 kJ when its temperature is increased by Δ T = 72 K , keeping its pressure constant. The value of C P C V for the gas is

Step-by-step solution

By the first law of thermodynamics Δ Q = Δ U + Δ W In an isobaric process Δ Q = C p Δ T and Δ U = C v Δ T (always) ∴ C p Δ T = C v Δ T + Δ W or Δ W = C p - C v Δ T or Δ W = R Δ T ∵ C p - C v = R ∴ Δ W = 8.3 × 7 2 = 597.6 J Δ Q = C p Δ T ∴ 1.6 × 1 0 0 0 = C p × 7 2 ⇒ C p = 1.6 × 1 0 0 0 7 2 = 22.2 J mol - 1 K - 1 Δ U = Δ Q = Δ W = 1.6 × 1 0 0