At 273 K temperature and 76 cm Hg pressure | vedclass

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(D) The ideal gas equation is given by $PV = nRT = \frac{w}{M} RT$. Rearranging for density $(d = \frac{w}{V})$, we get $d = \frac{PM}{RT}$. Thus, the

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$CO_2$

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(D) The ideal gas equation is given by $PV = nRT = \frac{w}{M} RT$. Rearranging for density $(d = \frac{w}{V})$, we get $d = \frac{PM}{RT}$. Thus, the molar mass $M = \frac{dRT}{P}$. Given: $d = 1.964 \ g \ L^{-1}$, $T = 273 \ K$, $P = 76 \ cm \ Hg = 1 \ atm$, and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$. Substituting the values: $M = \frac{1.964 	imes 0.0821 	imes 273}{1} \approx 44 \ g \ mol^{-1}$. The molar mass of $CO_2$ is $12 + 2 	imes 16 = 44 \ g \ mol^{-1}$. Therefore, the gas is $CO_2$.

At $273 \ K$ temperature and $76 \ cm \ Hg$ pressure, the density of a gas is $1.964 \ g \ L^{-1}$. The gas is

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