# Problem: A quantity of CO gas occupies a volume of 0.51 L at 1.5 atm and 312 K . The pressure of the gas is lowered and its temperature is raised until its volume is 3.1 L .Find the density of the CO under the new conditions.

###### FREE Expert Solution

We will calculate the density using the ideal gas law equation:

$\overline{){\mathbit{P}}{\mathbit{V}}{\mathbf{=}}{\mathbit{n}}{\mathbit{R}}{\mathbit{T}}}$

$\frac{\mathbf{P}\overline{)\mathbf{V}}}{\overline{)\mathbf{V}}}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{V}}\phantom{\rule{0ex}{0ex}}\mathbf{P}\mathbf{=}\frac{{\mathbf{n}}\mathbf{RT}}{\mathbf{V}}$

$\mathbf{P}\mathbf{=}\frac{{\mathbf{m}}\mathbf{RT}}{\mathbf{M}{\mathbf{V}}}$

For the new conditions, the only given value is the final volume of the CO gas. We’ll have to determine the final pressure and temperature.

$\overline{){\mathbf{P}}{\mathbf{V}}{\mathbf{=}}{\mathbf{n}}{\mathbf{R}}{\mathbf{T}}}$

Assuming the amount of gas in moles (n) remains constant and R is also a constant:

$\mathbf{PV}\mathbf{=}\overline{)\mathbf{nR}}\mathbf{T}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{PV}}{\mathbf{T}}\mathbf{=}\frac{\overline{)\mathbf{T}}}{\overline{)\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{PV}}{\mathbf{T}}\mathbf{=}\mathbf{1}$

Relate initial and final conditions:

$\overline{)\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}}}$

Given:

V1 = 0.51 L                V2 = 3.1 L
P1 = 1.5 atm              P2 = ?
T1 = 312 K                 T= ?

We can now calculate the density of CO gas in the new conditions:

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###### Problem Details

A quantity of CO gas occupies a volume of 0.51 L at 1.5 atm and 312 K . The pressure of the gas is lowered and its temperature is raised until its volume is 3.1 L .

Find the density of the CO under the new conditions.