Question

At what temperature will O2 have the same density
as N2 at 0°C, assuming pressure remains
constant?
(1) 39°C
(2) 39 K
(3) 273 K
(4) 312°C​

Answer 1

Answer:

a) 39° C

Explanation:

Let N2 be 1 and O2 be 2. Hence at 39 ℃, O2 will have same density as that of N2 at 0 ℃.

Answer 2

Answer:

OPTION (1) 39℃

Explanation:

$\sf We\:know\:that\:, \\ \\ \sf P = \frac{dRT}{M} \\ \\ \sf where, \: P \longrightarrow Pressure \\ \sf d \longrightarrow Density \\ \sf T \longrightarrow Temperature \:( in \: Kelvin ) \\\ \sf M \longrightarrow Molecular \:Mass \\ \\ \sf Rearranging \: the\:equation\:for\:d \\ \\ \sf d = \frac{PM}{RT} \\ \\ \sf Here\:P\:and\:R\:are\:constants \\ \\ \sf so\: d \: \alpha \: \frac{M}{T} \\ \\ \sf Hence \: {d}_{{O}_{2}} \: \alpha \: \frac{{M}_{{O}_{2}}}{{T}_{{O}_{2}}} \\ \\ \sf similarly \: {d}_{{N}_{2}} \: \alpha \: \frac{{M}_{{N}_{2}}}{{T}_{{N}_{2}}} \\ \\ \sf According\:to\:the\: question \: {d}_{{O}_{2}} = {d}_{{N}_{2}} \\ \\ \sf \longrightarrow \frac{{M}_{{O}_{2}}}{{T}_{{O}_{2}}} =\frac{{M}_{{N}_{2}}}{{T}_{{N}_{2}}} \\ \\ \sf Now\:putting\:the\:values \\ \\ \sf \longrightarrow \frac{32}{T} = \frac{28}{273} \\ \\ \sf \longrightarrow T = \frac{32×273}{28} \\ \\ \sf \longrightarrow T = 312 \:K \\ \\ \sf \longrightarrow \sf T = 312 - 273 = 39℃ \\ \\ \sf So\:the\: correct\:choice\:is\:option\:(1)$

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