Question

A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change in both temperature and pressure. find density of the gas

Answer 1

`\rho = 0.50 g/L`

Step-by-step explanation:

As we know that

PV = nRT

here we have

`P = 1.0 atm`

`P = 1.013 * 10^5 Pa`

so we have

`V = 1.4 * 10^(-3) m^3`

T = 290 K

now we have

`(1.013 * 10^5)(1.4 * 10^(-3)) = n(8.31)(290)`

`n = 0.06`

now the mass of gas is given as

`m = n M`

`m = (0.06)(28)`

`m = 1.65 g`

now density of gas when its volume is increased to 3.3 L

so we will have

`\rho = (m)/(V)`

`\rho = (1.65 g)/(3.3 L)`

`\rho = 0.50 g/L`