Question

If 528 kJ of heat are added to 2.25 kg of helium at constant volume process when the initial temperature is 40oC, find (a) the final temperature, (b) the change in enthalpy, (c) the change in entropy. 

Answer 1

(a) By the first law of thermodynamics:

$$Q=\Delta U+W=\Delta U+0=\Delta U,\\\space\\ Q=\frac 32\frac{m}{\Mu}R\Delta T,\\\space\\ \Delta T=\frac{2\Mu Q}{3mR}=75°\text{C},\\\space\\ T_f=T_i+\Delta T=40+75=115°\text{C}.$$

(b) The change in enthalpy equals the heat supplied, or 528 kJ.

(c) The change in entropy is

$$\Delta S=\frac{m}{\Mu}c_V\ln\frac{T_2}{T_1},\\\space\\ \Delta S=\frac{2.25}{4·10^{-3}}3116·\ln\frac{273+115}{273+40}=376494\text{ J}.$$