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# 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) $\dfrac{{10}}{{110}}$B) $\dfrac{{383}}{{283}}$C) $\dfrac{{110}}{{10}}$D) $\dfrac{{283}}{{383}}$

Last updated date: 02nd Aug 2024
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Hint: In this question, first we need to understand the molar form of the ideal gas equation. Then derive the expression of ratio of density and pressure. Then solve it using the inverse proportionality of these terms.

Complete step by step solution:
As we know that the molar form of ideal gas equation is $PV = nRT$, where $P$ is the pressure of the gas, $V$ is the volume of the gas, $R$ is the universal gas constant, $T$ is the absolute temperature, and $n$ is the number of moles of the gas, which can be written as $n = \dfrac{m}{M}$ denoted as ratio of the total mass of the gas $m$ and the molar mass of the gas $M$.
As we know that the ideal gas equation can also be written as,
$\rho = \dfrac{{PM}}{{RT}}$
Here, the density of the gas is $\rho$.
Now, we Assume the ratio of density and the pressure of the gas be $h$,
As we know the ideal gas equation as $\dfrac{\rho }{P} = \dfrac{M}{{RT}}$, the ratio can be expressed as $h = \dfrac{\rho }{P}$ or $h = \dfrac{M}{{RT}}$ so the ratio $h$ is inversely proportional to the absolute temperature $T$, that is, $h\alpha \dfrac{1}{T}$
Considering two ratios ${h_1}\;{\text{and}}\;{h_2}$ and two temperatures ${T_1}\;{\text{and}}\;{T_2}$, we get
$\dfrac{{{h_2}}}{{{h_1}}} = \dfrac{{{T_1}}}{{{T_2}}}$
Putting the given values in the above equation we get
$\dfrac{{{h_2}}}{{{h_1}}} = \dfrac{{\left( {273 + 10} \right)\;{\text{K}}}}{{\left( {273 + 110} \right)\;{\text{K}}}}$,
After simplification we get,
$\Rightarrow \dfrac{{{h_2}}}{{{h_1}}} = \dfrac{{283}}{{383}}$

Hence option (D) is correct.

Note:It is obvious that physical properties of the gases depend strongly on the conditions. We need a set of standard conditions so that the properties of gases can be properly compared to each other. While putting the value of temperature we must put the temperature in the Kelvin scale not in any other scale.