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 :

(1) Number of moles of first gas =n1NANumber of moles of second gas =n2NANumber of moles of third gas =n3NAIf there is no loss of energy thenP1V1+P2V2+P3V3=PV n1NART1+n2NART2+n3NART3=n1+n2+n3NARTmix ⇒Tmix=n1T1+n2T2+n3T3n1+n2+n3

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 :

(1) Number of moles of first gas =n1NANumber of moles of second gas =n2NANumber of moles of third gas =n3NAIf there is no loss of energy thenP1V1+P2V2+P3V3=PV n1NART1+n2NART2+n3NART3=n1+n2+n3NARTmix ⇒Tmix=n1T1+n2T2+n3T3n1+n2+n3

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Three perfect gases at absolute temperatures T₁ , T₂and T₃ are mixed. The masses of molecules are m₁, m₂and m₃ and the number of molecules are n₁ , n₂ and n₃ respectively. Assuming no loss of energy, the final temperature of the mixture is :

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{n_{1} {T_{1}}^{2} + n_{2} {T_{2}}^{2} + n_{3} {T_{3}}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{{n_{1}}^{2} {T_{1}}^{2} + {n_{2}}^{2} {T_{2}}^{2} + {n_{3}}^{2} {T_{3}}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{(T_{1} + T_{2} + T_{3})}{3}$

$\frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

(1) Number of moles of first gas = $\frac{n_{1}}{N_{A}}$

Number of moles of second gas = $\frac{n_{2}}{N_{A}}$

Number of moles of third gas = $\frac{n_{3}}{N_{A}}$

If there is no loss of energy then

$P_{1} V_{1} + P_{2} V_{2} + P_{3} V_{3} = P V \frac{n_{1}}{N_{A}} R T_{1} + \frac{n_{2}}{N_{A}} R T_{2} + \frac{n_{3}}{N_{A}} R T_{3} = \frac{n_{1} + n_{2} + n_{3}}{N_{A}} R T_{m i x} \Rightarrow T_{m i x} = \frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$