Answer to Question #121612 in Molecular Physics | Thermodynamics for Gopaul Yesuah

Question

A certain perfect gas is heated at constant pressure from 15 degrees celcius to 95 degrees celcius, the energy required is 1136 KJ / kg.
When this same gas is heated at constant volume to the same temperatures, the energy required is 808
KJ / kg. Calculate:
(i) Mass heat at constant pressure, cp
(ii) Mass heat at constant volume, CV
(iii) The value of γ
(iv) The particular gas constant, Rp
(v) Molar mass, Mr

Answer 1

Let the mass of gas $m$ be 1 kg, $M_r$be the molar mass, $\Delta T$ be the change of temperature.

In an isobaric process (see https://en.wikipedia.org/wiki/Isobaric_process)

$Q_1 = \dfrac{m}{M_r} C_p\Delta T,$

in an isochoric process

$Q_2 = \dfrac{m}{M_r} C_v\Delta T,$

where $C_p, \; C_v$ are the molar heat capacities. Therefore, $\gamma = \dfrac{C_p}{C_v} = \dfrac{Q_1}{Q_2} = \dfrac{1136\,\mathrm{kJ/kg}}{808\,\mathrm{kJ/kg} } \approx 1.41.$

But we should determine mass heat, so we should consider $c_p =\dfrac{C_p}{M_r}, \; c_v =\dfrac{C_v}{M_r}$ , so

$Q_1 = m c_p\Delta T \; \text{and} \; Q_2= m c_v\Delta T$ .

Therefore,

(i) $c_p = \dfrac{Q_1}{m\Delta T} = \dfrac{1136\,\mathrm{kJ}}{1\,\mathrm{kg}\cdot80\,\mathrm{K}} = 14.2\,\mathrm{kJ/kg/K}$ .

(ii) $c_v = \dfrac{Q_2}{m\Delta T} = \dfrac{808\,\mathrm{kJ}}{1\,\mathrm{kg}\cdot80\,\mathrm{K}} = 10.1\,\mathrm{kJ/kg/K}$

(iii) $\gamma \approx 1.41.$

(iv) $R_p = c_p-c_v = 14.2 \,\mathrm{kJ/kg/K} -10.1 \,\mathrm{kJ/kg/K} = 4.1\,\mathrm{kJ/kg/K}.$

(v) $M_r = \dfrac{R}{R_p} = \dfrac{8.31\,\mathrm{J/mol/K}}{4100\,\mathrm{J/kg/K}} \approx 0.002\,\mathrm{kg/mol} .$