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Q.

A glass vessel contains air at 60^oC. To what temperature must it be heated to expel one third of the air, the pressure remaining constant. (Neglect the expansion of the vessel)

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a

327^oC

b

226.5^oC

c

127^oC

d

427^oC

answer is B.

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Detailed Solution

$\frac{PV}{mT}$ = constant ;
Tm = constant; T₁m₁ = T₂m₂ $i f \frac{m}{3} i s e x p e l l e d t h e n m - \frac{m}{3} t h a t i s \frac{2 m}{3} i s r e m a i n e d$
$(333) (m_{1}) = T_{2} (\frac{2 m_{1}}{3})$
T₂ = 499.5 K $\Rightarrow$ t₂ = 226.5^oC

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