Question

A sample of 5.0 mol CO2(g) is originally confined in 15 dm3 at 280 K and then undergoes adiabatic expansion against a constant pressure of 78.5 kPa until the volume has increased by a factor of 4.0. Calculate q, w, delta T, and delta U, and delta H. (The final pressure of the gas is not necessarily 78.5 kPa.) If the enthalpy you calculate does not equal -4.5 kJ and the temperature change - 24 K, you made a mistake.

Answer 1

ym=0.5 I= 280 K イ T. 2- I S& 3 了 γ-1 1-31-I 3

Answer 2

For an adiabatic process q=0 Work =-Pex ×∆V 78.5×10^³×((60-15)×0.001)) =-3532.5J or -3.5325kJ ∆T=w/n(Cp-R)=-3532.5÷5(37.11-8.315) =-24.53K ∆U=q+w but q=0 ∆U=w =-3532.5J ∆H=∆U+nR∆T =-3532.5+(5×8.314×-24.53) =-4552.40J or -4.552kJ