At 273 K and 76 cm Hg pressure, the volume of 0.64 g of a gas is 224 mL. what temperature 1 g of this gas will occupy a volume of 1 litre at 1 atm pressure?

Explanation

Given P=76 cm Hg=1 atm, T=273 K,
V=224 mL=0.224 L
Number of moles `(n)=(0.64)/(M)` mol [molar mass of the gas `=M*g*mol^(-1)`]
Putting the values of P, V, n and T into the equation `PV=nRT` gives
`1atmxx0.224L=(0.64)/(M)xx0.0821L*atm*mol^(-1)*K^(-1)xx273K`
or, `M=64.04`
Therefore, number of moles of 1 g of gas `=(1)/(64.04)` mol
substituting P=1 atm, `V=1L,n=(1)/(64.04)` mol into equation pV=nRT, we obtain-
`T=(PV)/(nRT)=(1atmxx1L)/((1)/(64.04)molxx0.0821L*atm*mol^(-1)*K^(-1))=780.02K`
Therefore, at (780.02-273)=`507.02^(@)C` and 1 atm pressure the volume of 1 g of gas will be 1 L.