Question

A mixture of $10\text{g}$ of hydrogen and $40\text{g}$ of helium are kept in a closed vessel. To change the temperature of the mixture by ${50}^{\circ }\text{C}$, the amount of heat energy should be added to the mixture is
(Given $R=2{\text{cal mol}}^{-1}{}^{\circ }{\text{C}}^{-1}$)

• A. $2500\text{cal}$
• B. $2000\text{cal}$
• C. $2750\text{cal}$
• D. None of these

Answer 1

The correct option is C $2750\text{cal}$

Given,
Weight of hydrogen $=10\text{g}$
Weight of helium $=40\text{g}$
No. of moles of hydrogen $\left(n_1\right)=\frac{10}{2}=5$
No. of moles of helium $\left(n_2\right)=\frac{40}{4}=10$
Since, hydrogen is diatomic and helium is monoatomic, we get $(C_V{)}_1=\frac{5R}{2}$
$(C_V{)}_2=\frac{3R}{2}$
We know that,
$(C_V{)}_{\text{mixture}}=\frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}$
$=\frac{5\left(\frac{5R}{2}\right)+10\left(\frac{3R}{2}\right)}{15}$
$=\frac{11R}{6}$
$\therefore \Delta Q=n(C_V{)}_{mixture}\Delta T=\frac{15\times 11\times 2\times 50}{6}=2750\text{cal}$