Question

Calculate the no. of molecules present in 6oo cm³ of a gas at 99 kPa pressure at 25 degrees Celsius temperature? Given $R=8.314J{K}^{-1}mo{l}^{-1}$.
In this problem how 99 kPa is converted to 0.99 atm and why?

Answer 1

Given,

Volume = $V=600c{m}^3=600\times {10}^{-6}{m}^3$

Temperature = $T={25}^{o}C=273+25=298K$

Pressure = $P=99kPa=99000Pa$

From universal gas equation we have,

$PV=nRT$

$n=\frac{PV}{RT}=\frac{99000\times 600\times {10}^{-6}}{8.314\times 298}=0.23975$

Since 1 mole of a gas has $6.02\times {10}^{23}$ thus the number of molecules in the 0.23975 moles is given by,

$N=0.23975\times 6.02\times {10}^{23}=1.44\times {10}^{23}$

As we know
1 kilo-pascal = 0.00987 atmosphere
So, 99 kPa = 0.99 atm.