Question

A flask containing 2.0 moles of He gas at 1.0 atm and 300 K is connected to another flask contain \( N _{2}(g) \) at the same temperature and pressure by a narrow tube of negligible volume. Volume of nitrogen flask is three times volume of He -flask. Now the He -flask is placed in a thermostant at 2 K and \( N _{2} \)-flask in another thermostat at 400 K . What is the final pressure and final number of mo in each flask? a) 2.06 atm b) 1.06 atm c) 4.23 atm d) 3.26 atm

Answer 1

Correct option is b) 1.06 atm

We know,

\(n \propto \frac VT\)

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

\(\frac{n_{He}}{n_{N_2}} = \frac 13 \times \frac{400}{200} =\frac 23\)

Also,

\(n_{He} + n_{N_2} = 8\)

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

\(\Rightarrow n_{He} = \frac{16}5, n_{N_2} = \frac{24}5\)

Since volume of the Helium flask remains constant-

Also, \(\frac 1{300 \times 2} = \frac P{200 \times \frac{16}5}\)

\(\Rightarrow P = 1.06 atm\)