A mixture of $1 mol$ helium and $2 mol$ nitrogen (rigid diatomic) is heated at constant pressure from $300 K$ to $500 K$ . Treating the gases as ideal with $R = 8.314 J mo l^{- 1} K^{- 1}$ , the enthalpy change is nearest to

Answer: D

Δ H = (n_{He} C_{P, He} + n_{N_{2}} C_{P, N_{2}}) Δ T

with

C_{P} = 2.5 R

for He and

3.5 R

for

N X_{2}

. So

Δ H = (2.5 + 7.0) \times 8.314 \times 200 = 15797 J = 15.80 kJ

. The value

10.81 kJ

Δ U