The density of a gas at normal pressure and 2 | vedclass

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Question

(C) Given that the density of a gas $(d_1)$ at temperature $T_1 = 27^{\circ}C = 300 \ K$ is $24$.We need to find the density $(d_2)$ at temperature $T

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) Given that the density of a gas $(d_1)$ at temperature $T_1 = 27^{\circ}C = 300 \ K$ is $24$.We need to find the density $(d_2)$ at temperature $T_2 = 127^{\circ}C = 400 \ K$ while keeping the pressure $(P)$ constant.From the ideal gas equation,$PV = nRT = (m/M)RT$,we can write $P = (m/V)(RT/M) = d(RT/M)$.Thus,$d = \frac{PM}{RT}$.Since pressure $(P)$ and molar mass $(M)$ are constant,we have $d \propto \frac{1}{T}$.Therefore,$\frac{d_1}{d_2} = \frac{T_2}{T_1}$.Substituting the values: $\frac{24}{d_2} = \frac{400}{300}$.$d_2 = \frac{24 	imes 300}{400} = 18$.Thus,the density at $127^{\circ}C$ is $18$. The correct option is $(c)$.

The density of a gas at normal pressure and $27^{\circ}C$ temperature is $24$. Keeping the pressure constant,the density at $127^{\circ}C$ will be:

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