An ideal gas with $γ = C_{P} / C_{V} = 1.40$ absorbs $1.90 kJ$ of heat per mole at constant pressure. With $R = 8.314 J mo l^{- 1} K^{- 1}$ , the temperature rise of one mole is (in K, one decimal place)

Numerical answer type, answer in K, tolerance 0.1.

Answer: 65.3 K

From

C_{P} - C_{V} = R

and

C_{P} / C_{V} = γ

C_{P} = γ R / (γ - 1) = 1.40 (8.314) /0.40 = 29.099 J mo l^{- 1} K^{- 1}

Δ T = q / C_{P} = 1900/29.099 = 65.29 K