A flask containing same He gas at 1.5 ~atm \& 300 ~K is connected to another flask, four times larger in volume containing N_2 at same temperature and pressure by means of narrow negligible volume tube. The bigger flask

Correct answer: 2.2 aligned & n_1=x moles n_2=4 x moles & V_1 n_1 T_1 =(V_2)/(n_2 T_2) aligned (V)/(n_1 × 300)=(4 V)/(n_2 × 500)...(i) And n_2+n_1=x+4 x=5 x...(ii) From (i) \& (ii) n_1=(25 x)/(17) Then (P_1)/(P_2)=(n_1)/(n_2) in flask 1. (1.5)/(P)=(x)/(n_1) P=2.2 ~atm

JEE Advanced

Exam: JEE Advanced
Subject: Chemistry
Chapter: States of Matter

JEE Advanced · Chemistry · States of Matter

A flask containing same He gas at 1.5 ~atm \& 300 ~K is connected...

A flask containing same He gas at

1.5 \mathrm{~atm} \& 300 \mathrm{~K}

is connected to another flask, four times larger in volume containing

\mathrm{N}_2

at same temperature and pressure by means of narrow negligible volume tube. The bigger flask was then kept in a thermostat both maintained at 500 K , while other still at 300 K . Find final pressure. (mark the answer in atm) $A flask containing same He gas at 1.5 ~atm \& 300 ~K is connected to another flask, four times larger in volume$

Correct answer

2.2

Step-by-step solution

$A flask containing same He gas at 1.5 ~atm \& 300 ~K is connected to another flask, four times larger in volume$

\begin{aligned} & \mathrm{n}_1=\mathrm{x} \text { moles } \quad \mathrm{n}_2=4 \mathrm{x} \text { moles } \\ & \frac{V_1}{\mathrm{n}_1 T_1}=\frac{V_2}{n_2 T_2}\end{aligned}

\Rightarrow \frac{V}{n_1 \times 300}=\frac{4 V}{n_2 \times 500}...(i)

And

\mathrm{n}_2+\mathrm{n}_1=\mathrm{x}+4 \mathrm{x}=5 \mathrm{x}...(ii)

From (i) \& (ii)

n_1=\frac{25 x}{17}

Then

\frac{P_1}{P_2}=\frac{n_1}{n_2}

in flask 1.

\frac{1.5}{P}=\frac{x}{n_1} \Rightarrow P=2.2 \mathrm{~atm}

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