Air is filled at 60° C in a vessel of open mouth. The vessel is heated to a temperature T so that 1/4 textth part of air escapes. Assuming the volume of the vessel remaining constant, the value of T is

Question

Air is filled at 60° C in a vessel of open mouth. The vessel is heated to a temperature T so that 1/4 textth part of air escapes. Assuming the volume of the vessel remaining constant, the value of T is (A)  80° C  (B)  444° C  (C)  333° C  (D)  171° C

Tardigrade Admin · Accepted Answer

For open mouth vessel, pressure is constant. Volume is also given constant.
Hence from

p V = μ RT = (\frac{m}{M}) RT

\Rightarrow T \propto \frac{1}{m}

\Rightarrow \frac{T _{1}}{T _{2}} = \frac{m _{2}}{m _{1}}

∵ \frac{1}{4} t h

part escapes,
so remaining mass in the vessel

m_{2} = \frac{3}{4} m_{1}

\Rightarrow \frac{( 273 + 60 )}{T} = \frac{3/4 m _{1}}{m _{1}}

\Rightarrow T = 444 K = 17 1^{\circ} C

Q. Air is filled at $6 0^{\circ} C$ in a vessel of open mouth. The vessel is heated to a temperature $T$ so that $1/ 4^{th}$ part of air escapes. Assuming the volume of the vessel remaining constant, the value of $T$ is

$8 0^{\circ} C$

$44 4^{\circ} C$

$33 3^{\circ} C$

$17 1^{\circ} C$

Q. Air is filled at 60∘C in a vessel of open mouth. The vessel is heated to a temperature T so that 1/4th part of air escapes. Assuming the volume of the vessel remaining constant, the value of T is

80∘C

444∘C

333∘C

171∘C

For open mouth vessel, pressure is constant. Volume is also given constant. Hence from pV=μRT=(Mm​)RT ⇒T∝m1​ ⇒T2​T1​​=m1​m2​​ ∵41​th part escapes, so remaining mass in the vessel m2​=43​m1​ ⇒T(273+60)​=m1​3/4m1​​ ⇒T=444K=171∘C

Q. Air is filled at $6 0^{\circ} C$ in a vessel of open mouth. The vessel is heated to a temperature $T$ so that $1/ 4^{th}$ part of air escapes. Assuming the volume of the vessel remaining constant, the value of $T$ is

$8 0^{\circ} C$

$44 4^{\circ} C$

$33 3^{\circ} C$

$17 1^{\circ} C$