Two thermally insulated vessels 1 and 2 of volumes V₁ and V₂ are joined with a valve and filled with air at temperatures (T₁ , T₂) and pressures (P₁ , P₂) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium?
Answers
Answer:
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Explanation:
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Answer:
n1 = P1V1RT1n2 = P2V2RT2
Explanation:
There will be no change in number of moles if the bases are joined by the valves. Therefore, from gas equation
PV = nRT
P1V1/RT1 + P2V2/RT2 = P(V1+V2)/RT
≈ P1V1T2+P2V2T1/T2T2 = P(V1+V2)/T
T = P(V1+V2)T2T2/(P1V1T2+P2V2T)
Now, according to Boyle's law(pressure = constant) P1V1+P2V2 = P(V1+V2)
Hence, T = (P1V1+P2V2)T1T1/P1V1T1+P2V2T1
There will be no change in number of moles if the vessels are joined by the valves. Therefore, from gas equation
PV = nRTP1V1/RT1 + P2V2/RT2 = P(V1+V2)/RT
≈ P1V1T2+P2V2T1/T1T2 = P(V1+V2)/T
T = P(V1+V2)T1T2/P1V1T2+P2V2T
Now, according to Boyle's law(pressure = constant) P1V1+P2V2 = P(V1+V2)
Hence, T = (P1V1+P2V2)T1T1/P1V1T1+P2V2T1
n1 = P1V1RT1n2 = P2V2RT2