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A glass vessel contains air at 60 °C. 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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327 °C

226.5 °C

427 °C

127 °C

answer is B.

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Detailed Solution

$\frac{PV}{mT} = constant; Tm = constant; T_{1} m_{1} = T_{2} m_{2}$

$333 (m_{1}) = T_{2} (\frac{2 m_{1}}{3}) [∵ m_{2} = m_{1} - \frac{m_{1}}{3} = \frac{2 m_{1}}{3}]$

$T_{2} = 499 .5 K \Rightarrow T_{2} = 226 .5^{0} C$

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