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Question: A mixture of 4 gm Helium and 28 gm of nitrogen is enclosed in a vessel of constant volume of 300K. F...

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) .

Explanation

Solution

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×m×ΔtQ = C\times m\times \Delta t
Q = quantity of heat absorbed
M = mass of the body
Δt\Delta t = rise in temperature of the body

Complete step by step solution:
Given, T = 300K ; v’ = 2v
Root Mean Velocity
v=3RTMv = \sqrt{\dfrac{3RT}{M}}
Since, the constants are same therefore, no need to put them
vv=TT\Rightarrow \dfrac{v'}{v} = \sqrt{\dfrac{T'}{T}}
2vv=T300\Rightarrow \dfrac{2v}{v} = \sqrt{\dfrac{T'}{300}}
T’=1200K
Number of moles of Helium= n1=44=1{{n}_{1}}=\dfrac{4}{4}=1
Number of moles of Nitrogen=n2=2828=1{{n}_{2}}=\dfrac{28}{28}=1
Total number of moles
n=n1+n2{{n}_{1}}+{{n}_{2}} =1+1=2
For Helium, Cv1=32R{{C}_{v1}}=\dfrac{3}{2}R
For Nitrogen, Cv2=52R{{C}_{v2}}=\dfrac{5}{2}R

Specific heat capacity of mixture
Cv=n1Cv1+n2Cv2n1+n2{{C}_{v}}=\dfrac{{{n}_{1}}{{C}_{v1}}+{{n}_{2}}{{C}_{v2}}}{{{n}_{1}}+{{n}_{2}}}
Cv=1×32R+1×52R1+1=2R{{C}_{v}}=\dfrac{1\times \dfrac{3}{2}R+1\times \dfrac{5}{2}R}{1+1}=2R

Net heat absorbed at constant volume
Qv=nCv(TT){{Q}_{v}} = n{{C}_{v}}(T'-T)
Qv=2×2R(1200300)\Rightarrow {{Q}_{v}} = 2\times 2R(1200-300)
3600R\Rightarrow 3600R
After putting the values of Gas constant which is 8.314
3600×8.314\Rightarrow 3600\times 8.314
29930.4J\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 Cv{{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 Cp{{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.