Question

The ideal gas law is PV-nRT For R in units of 0.08206 Lit atm K-1.mol-1 determine the density of argon gas at 10.0 atm and 25.0°C in units of grams per cm3.

Answer 1

Answer: The density of argon gas is
`0.0163g/cm^3`

Step-by-step explanation:

To calculate the density of gas, we use the equation given by ideal gas equation:

`PV=nRT`

Number of moles (n) can be written as:
`n=(m)/(M)`

where, m = given mass

M = molar mass

`PV=(m)/(M)RT\\\\PM=(m)/(V)RT`

where,

`(m)/(V)=d` which is known as density of the gas

The relation becomes:

`PM=dRT` .....(1)

We are given:

M = molar mass of argon = 39.95 g/mol

R = Gas constant =
`0.08206\text{ L atm }mol^(-1)K^(-1)`

T = temperature of the gas =
`25^oC=[25+273]K=298K`

P = pressure of the gas = 10.0 atm

Putting values in equation 1, we get:

`10.0atm* 39.95g/mol=d* 0.08206\text{ L atm }mol^(-1)K^(-1)* 298K\\\\d=16.3g/L`

Converting the calculated density into
`g/cm^3`, we use the conversion factor:

`1L=1000cm^3`

Converting the density into
`g/cm^3`, we get:

`\Rightarrow (16.3g)/(L)* ((1L)/(1000cm^3))=0.0163g/cm^3`

Hence, the density of argon gas is
`0.0163g/cm^3`