Question

A flask containing 2.0 moles of $He$ gas at 1.0 atm and 300 K is connected to another flask containing $N_2\left(g\right)$ at the same temperature and pressure by a narrow tube of negligible volume.Volume of the nitrogen flask is three times volume of $He-$ flask. Now the $He-$flask is placed in a thermostat at 200 K and $N_2-$ flask in another thermostat at 400 K. What is the final pressure in the flask containing $He$?

• A. $2.06atm$
• B. $1.06atm$
• C. $4.23atm$
• D. $3.26atm$

Answer 1

Answer: (B)

$1.06atm$

We know,

$n\propto \frac{V}{T}$

Volume of the nitrogen flask is three times volume of He− flask

$\frac{n_{He}}{n_{N_2}}=\frac{1}{3}\times \frac{400}{200}=\frac{2}{3}$

Also,

$n_{He}+n_{N_2}=8$

Volume of the nitrogen flask is three times volume of He− flask

$⟹n_{He}=\frac{16}{5}$, $n_{N_2}=\frac{24}{5}$

Since volume of the Helium flask remains constant-

Also,$\frac{1}{300\times 2}=\frac{P}{200\times \frac{16}{5}}⟹P=1.06atm$