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2 years ago

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Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int

o a beaker. After the nitrogen evaporates, how much volume does it occupy if its density is equal to that of the dry air at sea level

1 answer:

neonofarm [45]2 years ago

3 0

Answer:

The value is $V_n = 2.2498 \ m^3$

Explanation:

From the question we are told that

The volume of liquid nitrogen is $V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3$

The density of nitrogen at gaseous form is $\rho_n = 1.2929 \ kg/m^3$ = The dry air at sea level

Generally the density of nitrogen at liquid form is

$\rho _l = 808 \ kg/m^3$

And this is mathematically represented as

$\rho_l = \frac{m}{V_l }$

=> $m = \rho_l * V_l$

Now the density of gaseous nitrogen is

$\rho_n = \frac{m}{V_n }$

=> $m = \rho_n * V_n$

Given that the mass is constant

$\rho_n * V_n = \rho_l * V_l$

$1.2929* V_n = 808 * 3.6*10^{-3}$

=> $V_n = 2.2498 \ m^3$

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