Question

A one litre gas bulb is evacuated and weighed, the weight is 500g. It is then filled with an ideal gas at 1atm pressure at 312.5 K. The weight of the filled bulb is 501.2 g. The molar weight of the gas is: ($R=8\times {{10}^{-2}}L\text{ }atm\text{ }{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}$ )
(A) 28
(B) 32
(C) 30
(D) 24

Answer 1

We can use the ideal gas equation and can easily find the weight of the gas present in the bulb. This equation is as below.
${PV=}{nRT}$
The formula that relates moles of gas with its weight is
$\text{Number of moles = }\dfrac{\text{weight}}{\text{molecular weight}}$

Complete step by step solution:
-We are given that the weight of the gas bulb is 500g when it is evacuated. When it is filled with an ideal gas the weight of bulb is 501.5g
So, the weight of gas present is 501.2g - 500g = 1.2g
-Using the ideal gas formula formula
${PV=}{nRT}$
Where n can also be written as
$\text{Number of moles (n)= }\dfrac{\text{weight}}{\text{molecular weight}}$
Here, we can write weight as W and Molecular weight as M.W.
So, we can write the ideal gas equation as
$\text{PV=}\dfrac{WRT}{(M.W)}$
Here, P is pressure and it is given 1atm, V volume and it is given 1 liter, therefore, putting the values of P, V, W, R and T in the equation we get,
$1\times 1=\dfrac{1.2\times 8\times {{10}^{-2}}\times 312.5}{M.W.}$
So,
$M.W.=1.2\times 8\times {{10}^{-2}}\times 312.5$
$M.W.=3000\times {{10}^{-2}}$
Hence,
$M.W.=30gmmo{{l}^{-1}}$

So, from the above we can conclude that option (C) is correct.

Note: Remember that here it is not required to convert the weight of gas into kilogram unit because we will describe molecular weight in $gmmo{{l}^{-1}}$ unit in which weight is expressed in grams, so both will cancel out each other. If we put the weight of gas in kg in the formula, then the answer would be incorrect as we will use $gmmo{{l}^{-1}}$ unit of molecular weight and both units will not cancel out each other.