Solution:
Since $\frac{\rho}{P} = \frac{M}{RT}$, the ratio is inversely proportional to the absolute temperature $T$. Thus, $x_2 = x_1 (\frac{T_1}{T_2}) = x \times \frac{10 + 273}{110 + 273} = x \times \frac{283}{383}$.
At $10^\circ\text{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $110^\circ\text{C}$ this ratio is (2008)
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