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 1

Answer: (D)

$\frac{283}{383} x$

Answer 2

Let the mass of the gas be m. At a fixed temperature and pressure, volume is fixed. Density of the gas, $\rho = \frac{m}{V}$ Now $\frac{\rho}{P} = \frac{m}{PV} = \frac{m}{nRT}$ $\Rightarrow \frac{m}{nRT} = x$ (By question) $\Rightarrow xT = \text{constant} \Rightarrow x_{1}T_{1} = x_{2} T_{2}$ $\Rightarrow x_{2} \Rightarrow \frac{x_{1}T_{1}}{T_{2}} = \frac{283}{383} x \begin{bmatrix}\therefore&&\\\ T_{1}&=&283K\\\T_{2}&=&383K\end{bmatrix}$