Question
At $${10^ \circ }C$$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $$x.$$ At $${110^ \circ }C$$ this ratio is
A.
$$x$$
B.
$$\frac{{383}}{{283}}x$$
C.
$$\frac{{10}}{{110}}x$$
D.
$$\frac{{283}}{{383}}x$$
Answer :
$$\frac{{283}}{{383}}x$$
Solution :
Concept
Use ideal gas equation to find the ratio between density of a fixed mass of an ideal gas and its pressure.
Ideal gas equation
$$\eqalign{
& pV = nRT \cr
& \frac{{pV}}{m} = \frac{1}{M}RT\left( {\because n = \frac{m}{M}} \right) \cr
& \frac{p}{\rho } = \frac{{RT}}{M}\,\,\left( {\because \frac{V}{m} = \frac{1}{\rho }} \right) \cr
& \therefore \frac{\rho }{p} \propto \frac{1}{T} \cr} $$
Molecular mass $$M$$ and universal gas constant $$R$$ remains same for a gas.
So, for two different situations i.e. at two different temperatures and densities
$$\eqalign{
& \therefore \frac{{\frac{{{\rho _1}}}{{{p_1}}}}}{{\frac{{{\rho _2}}}{{{p_2}}}}} = \frac{{{T_2}}}{{{T_1}}} \Rightarrow \frac{x}{{\left( {\frac{{{\rho _2}}}{{{p_2}}}} \right)}} = \frac{{383\,K}}{{283\,K}} \cr
& \Rightarrow \frac{{{\rho _2}}}{{{p_2}}} = \frac{{283}}{{383}}x \cr} $$