Question

The constant- pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression $C_{p} /(JK^{-1} )$ = 20.17 + 0.3665 (T/K). Calculate q,w, ΔU and ΔH when the temperature is raised from 25°C to 200°C
(a) at constant pressure
(b) at constant volume

Answer 1

A perfect gas, commonly known as an ideal gas, would be a gas that, in physical behavior, conforms towards the general gas law, which would be a specific, idealized relationship between pressure, volume, and temperature, and further calculation can be defined as follows:

For question a:

$q=\int C_p dT=\int^{200+273K}_{25+273 K} [20.17 + 0.3665 (\frac{T}{K})] dT JK^{-1}$

  $=[(20.17)T +\frac{1}{2}(0.3665)\times (\frac{T^2}{K})]^{473K}_{298K}\ JK^{-1}\\\\=[(20.17)\times(473-293) +\frac{1}{2}(0.3665)\times (473^2-293^2)]\ J\\\\=[(20.17)\times(180) +\frac{1}{2}(0.3665)\times (223729-85849)]\ J\\\\=[3630.6 +(0.18325)\times (137880)]\ J\\\\=[3630.6 +25266.51]\ J\\\\=[28897.11]\ J\\\\=[28.8 \times 10^{3}] \ J$

$W= -p \Delta V= -nR\Delta T= -(1.00\ mol) \times 8.3145\ J K^{-1}\ mol^{-1} \times 100\ K=-831\ J\\\\\Delta U= q+w=(28.8 -0.831)\ KJ= 27.969\ KJ$

 For question b:

In this question, the energy and enthalpy of a perfect gas depend on temperature alone, therefore $\Delta H= 28.8 \ KJ$ and $\Delta U= 27.969\ KJ$ as above at constant volume $w=0$ and $\Delta U=q$ so, $q=27.969 \ KJ$.

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