At 0,^{circ}text{C},the value of the density | vedclass

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(C) From the ideal gas equation,$PV = nRT = \frac{m}{M_w}RT$,where $m$ is the mass and $M_w$ is the molar mass.Rearranging for density $\rho = \frac{m

Vedclass · Accepted Answer

$\frac{273}{373}x$

Vedclass · Answer

(C) From the ideal gas equation,$PV = nRT = \frac{m}{M_w}RT$,where $m$ is the mass and $M_w$ is the molar mass.Rearranging for density $ho = \frac{m}{V}$,we get $P = \frac{ho RT}{M_w}$,which implies $\frac{ho}{P} = \frac{M_w}{RT}$.Given that at $T_1 = 273\,K$ $(0\,^{\circ}	ext{C})$,$\frac{ho_1}{P_1} = x = \frac{M_w}{R(273)}$.At $T_2 = 373\,K$ $(100\,^{\circ}	ext{C})$,the new quotient is $x' = \frac{M_w}{R(373)}$.Dividing the two expressions: $\frac{x'}{x} = \frac{M_w / (R \cdot 373)}{M_w / (R \cdot 273)} = \frac{273}{373}$.Therefore,$x' = \frac{273}{373}x$.

At $0\,^{\circ}\text{C}$,the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $100\,^{\circ}\text{C}$,this quotient is:

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