Get Answers to all your Questions

header-bg qa

 An open vessel at 27^{\circ} \mathrm{C} is heated until two fifth of the ain in it has escaped from the vessel. Assuming that the volume of the Vessel remains constant, the temperature at which the vessel has been heard in -
 

Option: 1

700 \mathrm{~K}


Option: 2

500 \mathrm{k}


Option: 3

750 \mathrm{k}


Option: 4

500^{\circ} \mathrm{C}


Answers (1)

best_answer


\mathrm{\text { given, temperature }\left(T_1\right)=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K} }
\begin{aligned} & \text { volume of vessel }=\text { constant } \\ & \text { press cure in vessel }=\text { constant } \end{aligned}

volume of ain reduced by \frac{2}{5} so the remaining volume of ain is \frac{3}{5}. Let, at \mathrm{T_1 }the volume of airs inside the vessel is n so at \mathrm{T_2}the volume of air will be \mathrm{\frac{3}{5} }n.
Now, a s p and vane constant, 10

\mathrm{n \cdot T_1=\frac{3}{5} n T_2-(1) }
Putting the value of \mathrm{T_1 }in equation (1) we gel,

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

 

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE