Question

A container holds 15.0 g of phosphorous gas at a pressure of 2.0 atm and a temperature of 20.0 Celsius. What is the density of the gas?
2.57 g/L
12.3 g/L
180 g/L
0.40 g/L

Answer 1

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We calculate as follows:

PV = nRT
PV = mRT/ Molar mass
m/V = P(molar mass)/RT
Density = P(molar mass)/RT 
Density = 2.0 ( 30.97 ) / 0.08206 ( 20 + 273.15) = 2.57 g/L <----First option

Answer 2

Answer: Density of the gas is 2.57 g/L.

Explanation: Converting the mass of phosphorous gas into its moles using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of phosphorous gas = 15 g

Molar mass of phosphorous gas = 31 g/mol

Using above equation, we get

$n=\frac{15g}{31g/mol}=0.484moles$

Using Ideal gas equation, which is:

$PV=nRT$

Given:

P = 2 atm

$T=20^oC=(273+20)K=293K$    (Conversion Factor: $0^oC=273K$ )

n = 0.484 moles (Calculated above)

$R=0.082057\text{ L atm }mol^{-1}K^{-1}$  (Gas Constant)

Putting all the values in above equation, we calculate the number of moles.

$2atm\times V=(0.484moles)(0.082057\text{ L atm }mol^{-1}K^{-1})(293K)$

V = 5.818 L

Now, to calculate density, we use the formula:

$Density=\frac{Mass}{Volume}$

$Density=\frac{15g}{5.818L}$

Density = 2.57 g/L