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At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $110^{\circ} \mathrm{C}$ this ratio is
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The correct answer is:
$\frac{283}{383} x$
Let the ratio of density and pressure be r.
From Ideal Gas Equation, \(\rho=\frac{\mathrm{PM}}{\mathrm{RT}}\)
\(\begin{aligned}
& \mathrm{r}=\frac{\rho}{\mathrm{P}}=\frac{\mathrm{M}}{\mathrm{RT}} \\
& \therefore \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{273+10}{273+110}=\frac{283}{383} \\
& \mathrm{r}_2=\frac{283}{383} \quad\left(\text { as, } \mathrm{r}_1=\mathrm{x}\right)
\end{aligned}\)
From Ideal Gas Equation, \(\rho=\frac{\mathrm{PM}}{\mathrm{RT}}\)
\(\begin{aligned}
& \mathrm{r}=\frac{\rho}{\mathrm{P}}=\frac{\mathrm{M}}{\mathrm{RT}} \\
& \therefore \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{273+10}{273+110}=\frac{283}{383} \\
& \mathrm{r}_2=\frac{283}{383} \quad\left(\text { as, } \mathrm{r}_1=\mathrm{x}\right)
\end{aligned}\)
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