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

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

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

$\rho = \frac{m}{V}$

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$