Question

A mixture of 4 gm Helium and 28 gm of nitrogen is enclosed in a vessel of constant volume of 300K. Find the quantity of heat absorbed by the mixture to double the root mean square velocity of its molecules. (R = Universal gas constant) .

Answer 1

Hint: Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 kilogram of substance by 1 Kelvin.
$Q = C\times m\times \Delta t$
Q = quantity of heat absorbed
M = mass of the body
$\Delta t$ = rise in temperature of the body

Complete step by step solution:
Given, T = 300K ; v’ = 2v
Root Mean Velocity
$v = \sqrt{\dfrac{3RT}{M}}$
Since, the constants are same therefore, no need to put them
$\Rightarrow \dfrac{v'}{v} = \sqrt{\dfrac{T'}{T}}$
$\Rightarrow \dfrac{2v}{v} = \sqrt{\dfrac{T'}{300}}$
T’=1200K
Number of moles of Helium= ${{n}_{1}}=\dfrac{4}{4}=1$
Number of moles of Nitrogen=${{n}_{2}}=\dfrac{28}{28}=1$
Total number of moles
n=${{n}_{1}}+{{n}_{2}}$ =1+1=2
For Helium, ${{C}_{v1}}=\dfrac{3}{2}R$
For Nitrogen, ${{C}_{v2}}=\dfrac{5}{2}R$

Specific heat capacity of mixture
${{C}_{v}}=\dfrac{{{n}_{1}}{{C}_{v1}}+{{n}_{2}}{{C}_{v2}}}{{{n}_{1}}+{{n}_{2}}}$
${{C}_{v}}=\dfrac{1\times \dfrac{3}{2}R+1\times \dfrac{5}{2}R}{1+1}=2R$

Net heat absorbed at constant volume
${{Q}_{v}} = n{{C}_{v}}(T'-T)$
$\Rightarrow {{Q}_{v}} = 2\times 2R(1200-300)$
$\Rightarrow 3600R$
After putting the values of Gas constant which is 8.314
$\Rightarrow 3600\times 8.314$
$\Rightarrow 29930.4J$
Hence, the answer of this question is 29930.4 J

Note: When the volume of solid remains constant after a small change in temperature, it is known as specific heat at constant volume. It is donated by ${{C}_{v}}$. When the pressure of solid remains constant when through a small range of temperature, it is known as specific heat at constant pressure. It is donated by ${{C}_{p}}$. The most important point to be noted in this question is to remember the units of specific heats. As seen above, we have not put the value of R because it would have complicated the question. Therefore, try to visualise the calculations first.