Question

A mixture of 4gm helium and 28gm nitrogen is enclosed in a vessel of constant volume 300K
Find the quantity of heat absorbed by the mixture to doubled the root mean velocity of its molecules

Answer 1

Answer: $\Delta Q = 29930.4\ J$

Explanation:

Given that,

Temperature = 300 K

Mass of helium $M_{He} = 4 gm$

Number of mole of helium = 1 mol

Mass of nitrogen $M_{N_{2}} = 28 gm$

Number of mole of nitrogen = 1 mol

Degree of freedom of helium f = 3

Degree of freedom of nitrogen f = 5

Specific heat of helium $c_{v} = \dfrac{3}{2}$

Specific heat of nitrogen $c_{v} = \dfrac{5}{2}$

Specific heat of mixture

$C_{v}_{mix} = \dfrac{n_{He}C_{v}+ n_{N_{2}C_{v}}}{n_{He}+n_{N_{2}}}$

$C_{v} = \dfrac{\dfrac{3}{2}R+ \dfrac{5}{2}R}{1+1}$

$C_{v} = 2 R$

We know that,

$v_{rms} = \sqrt{\dfrac{3RT}{M}}$

When the mixture to doubled the root mean velocity of its molecules then, the temperature is four times of the original temperature.

So, $T_{f} = 4 T_{i}$

$T_{f} = 4\times 300\ K$

$\Delta T = T_{f} - T_{i}$

$\Delta T = 1200K - 300 K$

$\Delta T= 900\ K$

Now, the heat is

$\Delta Q = nC_{v}\Delta T$

$\Delta Q = (n_{He}+n_{N_{2}})C_{v}\Delta T$

$\Delta Q = 2\times 2\times 8.314\times 900\ K$

$\Delta Q =29930.4\ J$

Hence, this is the required solution.