devarshi-dt-logo

Question:

Three perfect gases at absolute temperatures T1, T2, and T3 are mixed. The masses of molecules are m1, m2, and m3, and the number of molecules are n1, n2, and n3 respectively. Assuming no loss of energy, the final temperature of the mixture is:

(T1+T2+T3)/3

(n1T1+n2T2+n3T3)/(n1+n2+n3)

(n1T1^2+n2T2^2+n3T3^2)/(n1T1+n2T2+n3T3)

(n1T1^2+n2T2^2+n3T3^2)/(n1T1+n2T2+n3T3)

Solution:

Number of moles of first gas = n1NA
Number of moles of second gas = n2NA
Number of moles of third gas = n3NA
If no loss of energy then
P1V1 + P2V2 + P3V3 = PV
Using the ideal gas equation PV = nRT, we have:
n1RT1 + n2RT2 + n3RT3 = (n1 + n2 + n3)RT
Since R is a constant, we get:
n1T1 + n2T2 + n3T3 = (n1 + n2 + n3)T
T = (n1T1 + n2T2 + n3T3) / (n1 + n2 + n3)