Question

Which sample would have the same number of molecules as 11.2L of He (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.