Heat and Thermodynamics 4 Question 2
2. Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume?
[Take, $R=8.3 J / mol-K$ ]
(2019 Main, 12 April I)
(a) $19.7 J / mol-K$
(b) $15.7 J / mol-K$
(c) $17.4 J / mol-K$
(d) $21.6 J / mol-K$
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Solution:
- Let molar specific heat of the mixture is $C _V$.
Total number of molecules in the mixture
$$ =3+2=5 $$
$\therefore C _V$ can be determined using
$$ \begin{aligned} n C _V d T & =n _1 C _{V _1} d T+n _2 C _{V _2} d T \\ \text { or }\left(n _1+n _2\right)\left(C _V\right) _{\text {mix }} & =n _1 C _{V _1}+n _2 C _{V _2} \\ & {\left[\text { here, } n=n _1+n _2\right] } \end{aligned} $$
Here, $C _{V _1}=\frac{3 R}{2}$ (for helium); $n _1=2$
$$ C _{V _2}=\frac{5 R}{2} \text { (for hydrogen); } n _2=3 $$
[For monoatomic gases, $C _V=\frac{3}{2} R$ and for diatomic gases, $\left.C _V=\frac{5}{2} R\right]$
$$ \begin{array}{rlrl} & \therefore & 5 \times C _V & =\left(2 \times \frac{3 R}{2}\right)+\left(3 \times \frac{5 R}{2}\right) \\ \Rightarrow & 5 C _V & =\frac{21 R}{2} \\ \text { or } & C _V & =\frac{21 R}{10} \\ & =\frac{21 \times 8.3}{10}=\frac{174.3}{10} \end{array} $$
or
$$ C _V=17.4 J / mol-K $$
