Q104 / 200 · NEET 2008 (Screening) · Physics

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

  • A
    $x$
  • B
    $\frac{383}{283} x$
  • C
    $\frac{10}{110} x$
  • Correct answer
    D
    $\frac{283}{383} x$

Solution

Let the ratio of density and pressure be r. From Ideal Gas Equation, \(\rho=\frac{\mathrm{PM}}{\mathrm{RT}}\) \(\begin{aligned} & \mathrm{r}=\frac{\rho}{\mathrm{P}}=\frac{\mathrm{M}}{\mathrm{RT}} \\ & \therefore \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{273+10}{273+110}=\frac{283}{383} \\ & \mathrm{r}_2=\frac{283}{383} \quad\left(\text { as, } \mathrm{r}_1=\mathrm{x}\right) \end{aligned}\)

Hence, the correct answer is option: 'D'.

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