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An open vessel at \mathrm{27^{\circ}C}  is heated until two fifth of the air in it has escaped from the vessel,Assuming that the volume of the vessel remaining constant,the temperatue at which the vessel has been heated is ____

Option: 1

700 k


Option: 2

500 k


Option: 3

750 k


Option: 4

500o C


Answers (1)

best_answer

Given ,temperature \mathrm{\left(T_1\right)=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}}

Volume of vessel = constnt

pressure in vessel=constant

Volume of air reduced by \mathrm{\frac{2}{5}} so the remaining volume of air is \mathrm{\frac{3}{5}}.

Let at \mathrm{T_{1}} the volume of air inside the vessel is n 30 at \mathrm{T_{1}} the volume of air will be \mathrm{\frac{3}{5}n}

Now, as P and V are constnt , so

\mathrm{n \cdot T_1=\frac{3}{5} n T_2} ------------(1)

Putting the value of \mathrm{T_1} in equation (1) we get,

\mathrm{\begin{aligned} & n=300=\frac{3}{5} n \cdot T_2 \\ & \Rightarrow T_2=300 \times \frac{5}{3}=500 \mathrm{k} \end{aligned}}

 

Posted by

Divya Prakash Singh

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