Solution:
From ideal gas law, \(P = \frac{\rho RT}{M} \Rightarrow \frac{\rho}{P} = \frac{M}{RT}\). Hence, \(\frac{\rho}{P} \propto \frac{1}{T}\). Thus, the ratio becomes \[X \times \frac{273 + 10}{273 + 110} = X \frac{283}{383}\].
At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is X. At 110°C this ratio is
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