Question

How much heat energy should be added to the gaseous mixture consisting of one gram of hydrogen and one gram helium to raise its temperature from $0^{o}C$ to ${50}^{o}C$.
(a) at constant volume and
(b) at constant pressure, assuming both the gases to be ideal
$R=2cal/mol-K$, ${\gamma }_{H_2}=1.41$, ${\gamma }_{H_e}=1.67$

Answer 1

$cp-cv=R⟶\left(1\right)$ $\frac{cp}{cv}=r⟶\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$

$\Rightarrow cp-cv=R$

$rcv-cv=R$

$cv=\frac{R}{r-1}$

$\left(a\right).$ $cv$ for Hydrogen $=\frac{2}{1.41-1}=4.87cal/molek$

$cv$ for Helium $=\frac{2}{1.67-1}=2.98cal/molek$

$\Rightarrow Heat=n_{1}c{v}_{1}Q+n_{2}c{v}_{2}Q$

$=\frac{1}{2}\times 4.87\times 50+\frac{1}{4}\times 2.98\times 50$

$=121.75cal+37.25cal$

$=159cal$

$\left(b\right).$ $cp$ for hydrogen -

$cp=rcv$

$=1.41\times 4.87$

$=6.867cal/molek$

$cp$ for Helium

$\Rightarrow cp=rcv$

$=1.67\times 2.98cal/molek$

$=4.97cal/molek$

$\Rightarrow H=n_{1}c{p}_{1}Q+n_{2}c{p}_{2}Q$

$=\frac{1}{2}\times 6.87\times 50+\frac{1}{4}\times 4.97\times 50$

$=171.75cal+62.125cal$

$=233.875cal$

Hence, the answer is $233.875cal.$