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
(b) at constant pressure, assuming both the gases to be ideal
$R=2cal/mol-K,{\gamma }_{H_2}=1.41,{\gamma }_{He}=1.67$

Answer 1

According to the problem, for hydrogen,

$\frac{C_P}{C_V}=\gamma =1.41$ and $C_P-C_V=R=2$

$\therefore C_P=1.41C_V$ and $1.41C_V-C_V=2$ or $C_V=\frac{2}{0.41}=4.88$

$\therefore C_P=1.41\times 4.88=6.88$

For Helium,

$\frac{C_P}{C_V}=1.67$ and $C_P-C_V=2$

Again, $C_P=1.67C_V$ and $1.67C_V-C_V=2$ or, $C_V=\frac{2}{0.67}=2.985$

$\therefore C_P=1.67\times 2.985=4.98$

If $m$ be the mass of the gas and $M$ the molecular weight, then

the mass of the gas in mole will be $\frac{m}{M}$

(i) To raise the temperature of $\frac{m}{M}$ moles of gas by $\Delta T$ at constant volume,the energy added to the gas = $\left(\frac{m}{M}\right)\times C_V\times \Delta T$

Energy added to hydrogen = $\frac{1}{2}\times 4.88\times (50-0)=122cal$

Energy added to helium=$\frac{1}{4}\times 2.985\times (50-0)=37.3cal$

Therefore, energy given to the mixture at constant volume =122+37.3=159.3 cal.

(ii) The energy added to the gaseous mixture at constant pressure is again obtained by calculating the energy added to the individual gas.

The energy given to a gas=$\left(\frac{m}{M}\right)\times C_P\times \Delta T$

Energy added to hydrogen = $\frac{1}{2}\times 6.88\times (50-0)=172cal$

Energy added to helium=$\frac{1}{4}\times 4.98\times (50-0)=62.25cal$

Therefore, energy given to the mixture at constant volume =172+62.25=234.25 cal.