Question

A flask of 2 dm3 (dm cube) capacity contains O2 at 101.325 kPa and 300 K. The gas pressure is reduced to 0.1 Pa. Assuming ideal behaviour, answer the following:
a) what will be the volume of the gas which is left behind?
b) what amount of O2 and the corresponding number of molecules are left behind in the flask?
c) if now 2 g of N2 is introduced, what will be the pressure of the flask?

Answer 1

Dear student,
Given :P=101.325kPa =101325 Pa=1 atm
V= 2 dm³ =2 L
T=300 K
so number of moles n= $\frac{PV}{RT}$= $\frac{1\times 2}{0.0821\times 300}$=0.081 moles
1) Volume of gas left behind when presure is reduced to 0.1 pa
P= 0.1 pa = 9.8$\times {10}^{-7}atm$
V= $\frac{nRT}{P}$= $\frac{0.082\times 0.0821\times 300}{9.8\times {10}^{-7}}$=2.03$\times$${10}^7$ L
2)22.4 L O₂= $32g{O}_2$
so 2.03 $\times {10}^7$ L = 2.9$\times {10}^7$ gm
Number of particles
32 gm = $6.023\times {10}^{23}$
$2.9\times {10}^7=\frac{6.023\times {10}^{23}}{32}\times 2.9\times {10}^7=5.4\times {10}^{29}$ particles
Regards!