Question

Which sample would have the same number of molecules as 11.2L of He (g) at 273K and 202kPa?
1) 11.2L of N2(g) at 300K and 202kPa
2) 22.4L of Ne(g) at 546K and 404K
3) 11.2L of CH4(g) at 273K and 202kPa

Answer 1

Answer: 11.2 L of $CH_4(g)$ at 273K and 202kPa

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 202 kPa = 1.99 atm  ( 1kPa= 0.0098 atm)

V = Volume of gas = 11.2 L

n = number of moles = ?

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$273K$

$n=\frac{PV}{RT}$

$n=\frac{1.99\times 11.2}{0.0821 L atm/K mol\times 273K}=0.99moles$

According to avogadro's law, equal number of moles occupy equal volumes and contain equal number of molecules at same temperature and pressure conditions.

As 11.2 L of $CH_4(g)$ at 273K and 202kPa will have same moles as 11.2L of He (g) at 273K and 202kPa, thus they have same number of molecules.