A mixture of 10 g of hydrogen and 40 g of helium are kept in a closed vessel- To change the temperature of the mixture by 50- C- the amount of heat energy should be added to the mixture isGiven R-2 cal mol-1 -C-1

Given, Weight of hydrogen =10 g Weight of helium =40 g No. of moles of hydrogen (n_1)=102=5 No. of moles of helium (n_2)=404=10 Since, hydrogen is diatomic and ...

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Standard XIII

Physics

Equipartition Theorem

A mixture of ...

Question

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

2500 cal

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2000 cal

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2750 cal

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None of these

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Solution

The correct option is C $2750 cal$
Given,
Weight of hydrogen $= 10 g$
Weight of helium $= 40 g$
No. of moles of hydrogen $(n_{1}) = \frac{10}{2} = 5$
No. of moles of helium $(n_{2}) = \frac{40}{4} = 10$
Since, hydrogen is diatomic and helium is monoatomic, we get $(C_{V})_{1} = \frac{5 R}{2}$
$(C_{V})_{2} = \frac{3 R}{2}$
We know that,
$(C_{V})_{mixture} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$
$= \frac{5 (\frac{5 R}{2}) + 10 (\frac{3 R}{2})}{15}$
$= \frac{11 R}{6}$
$∴ Δ Q = n (C_{V})_{m i x t u r e} Δ T = \frac{15 \times 11 \times 2 \times 50}{6} = 2750 cal$

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

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