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Question

Chemistry Question on Ideal gas equation

Three perfect gases at absolute temperatures T1,T2T_{1}, T_{2} and T3T_{3} are mixed. The masses of molecules are m1m_{1}, m2m_{2} and m3m_{3} and the number of molecules are n1,n2n_{1}, n_{2} and n3n_{3} respectively. Assuming no loss of energy, the final temperature of the mixture is

A

n12T12+n22T22+n32T32n1T1+n2T2+n3T3\frac{n_{1}^{2} T_{1}^{2}+n_{2}^{2} T_{2}^{2}+n_{3}^{2} T_{3}^{2}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}

B

T1+T2+T33\frac{T_{1}+T_{2}+T_{3}}{3}

C

n1T1+n2T2+n3T3n1+n2+n3\frac{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}{n_{1}+n_{2}+n_{3}}

D

n1T12+n2T22+n3T32n1T1+n2T2+n3T3\frac{n_{1} T_{1}^{2}+n_{2} T_{2}^{2}+n_{3} T_{3}^{2}}{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}

Answer

n1T1+n2T2+n3T3n1+n2+n3\frac{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}{n_{1}+n_{2}+n_{3}}

Explanation

Solution

U=n1NT1+n2NT2+n3NT3n1N+n2N+n3NU =\frac{\frac{n_{1}}{N} T_{1}+\frac{n_{2}}{N} T_{2}+\frac{n_{3}}{N} T_{3}}{\frac{n_{1}}{N}+\frac{n_{2}}{N}+\frac{n_{3}}{N}} =n1T1+n2T2+n3T3n1+n2+n3=\frac{n_{1} T_{1}+n_{2} T_{2}+n_{3} T_{3}}{n_{1}+n_{2}+n_{3}}