Question

A quantity of CO gas occupies a volume of 0.64 L at 0.90 atm and 307 K . The pressure of the gas is lowered and its temperature is raised until its volume is 3.0 L .

Answer 1

This is an incomplete question, here is a complete question.

A quantity of CO gas occupies a volume of 0.48 L at 1.6 atm and 282 K. The pressure of the gas is lowered and its temperature is raised until its volume is 2.1 L. Find the density of the CO under the new conditions?

Answer : The density of the CO under the new conditions is, 0.213 g/L

Explanation :

First we have to determine the moles of CO gas by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 0.90 atm

V = volume of gas = 0.64 L

T = temperature of gas = 3 07 K

R = gas constant = 0.0821 L.atm/mol.K

n = number of moles of gas = ?

Now put all the given values in the above formula, we get:

$(0.90atm)\times (0.64L)=n\times (0.0821L.atm/mol.K)\times (307K)$

$n=0.0228mol$

Now we have to calculate the mass of CO gas.

$\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO$

Molar mass of $CO$ = 28g/mole

$\text{ Mass of }CO=(0.0228moles)\times (28g/mole)=0.638g$

Now we have to calculate the density of CO gas.

$\text{Density of CO gas}=\frac{\text{Mass of CO gas}}{\text{Volume of CO gas}}$

$\text{Density of CO gas}=\frac{0.638g}{3.0L}=0.213g/L$

Therefore, the density of the CO under the new conditions is, 0.213 g/L