Question

How much heat energy should be added to the gaseous mixture consisting of $1g$ of hydrogen and $1g$ of helium to raise its temperature from $0^{\circ }C$ to ${100}^{\circ }C$
at constant volume, $(R=2cal/molK)$?

• A. ${\left(\Delta Q\right)}_v=325.567cal$
• B. ${\left(\Delta Q\right)}_v=650cal$
• C. ${\left(\Delta Q\right)}_v=325cal$
• D. ${\left(\Delta Q\right)}_v=375cal$

Answer 1

The correct option is C ${\left(\Delta Q\right)}_v=325cal$

As hydrogen is diatomic and has molecular weight 2,
${\left(C_v\right)}_H=\frac{5}{2}R$
and ${\mu }_H=\frac{1}{2}$
While $He$ is monatomic and has molecular weight 4,
${\left(C_v\right)}_{He}=\frac{3}{2}R$ and ${\mu }_{He}=\frac{1}{4}$
So, by conservation of energy, we get
${\left(C_v\right)}_{mix}=\frac{\mu {\left(C_v\right)}_1+{\mu }_2{\left(C_v\right)}_2}{{\mu }_1+{\mu }_2}=\frac{\frac{1}{2}\times \frac{5}{2}R+\frac{1}{4}\times \frac{3}{2}R}{\frac{1}{2}+\frac{1}{4}}$
${\left(C_v\right)}_{mix}=\frac{13}{8}\times \frac{4}{3}R=\frac{13}{6}R$
So,
${\left(\Delta Q\right)}_v=\mu C_v\Delta T=(\frac{1}{2}+\frac{1}{4})\times \frac{13}{6}\times 2\times (100-0)=325cal$