At 10^{circ} C,the value of the density of a | vedclass

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Question

(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.Density $\rho = \frac{m}{V} = \frac{PM}{RT}

Vedclass · Accepted Answer

$\left( \frac{283}{383} \right)x$

Vedclass · Answer

(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.Density $ho = \frac{m}{V} = \frac{PM}{RT}$.The ratio of density to pressure is $\frac{ho}{P} = \frac{M}{RT}$.Since $M$ and $R$ are constants,$\frac{ho}{P} \propto \frac{1}{T}$.Let the ratio at $T_1 = 10^{\circ} C = 283 \ K$ be $x_1 = x$.Let the ratio at $T_2 = 110^{\circ} C = 383 \ K$ be $x_2$.Since $x \cdot T = 	ext{constant}$,we have $x_1 T_1 = x_2 T_2$.$x \cdot 283 = x_2 \cdot 383$.Therefore,$x_2 = \left( \frac{283}{383} ight)x$.

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:

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