Question
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) .
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×Δt
Q = quantity of heat absorbed
M = mass of the body
Δt = rise in temperature of the body
Complete step by step solution:
Given, T = 300K ; v’ = 2v
Root Mean Velocity
v=M3RT
Since, the constants are same therefore, no need to put them
⇒vv′=TT′
⇒v2v=300T′
T’=1200K
Number of moles of Helium= n1=44=1
Number of moles of Nitrogen=n2=2828=1
Total number of moles
n=n1+n2 =1+1=2
For Helium, Cv1=23R
For Nitrogen, Cv2=25R
Specific heat capacity of mixture
Cv=n1+n2n1Cv1+n2Cv2
Cv=1+11×23R+1×25R=2R
Net heat absorbed at constant volume
Qv=nCv(T′−T)
⇒Qv=2×2R(1200−300)
⇒3600R
After putting the values of Gas constant which is 8.314
⇒3600×8.314
⇒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. 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. 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.