Question

A tank contains 0.568 mol of molecular nitrogen (N2). Determine the mass (in grams) of nitrogen that must be removed from the tank in order to lower the pressure from 44.7 to 22.8 atm. Assume that the volume and temperature of the nitrogen in the tank do not change.

Answer 1

Answer:

$\Delta M = 7.812 gm$

Explanation:

we know that for ideal gag we have

pV =nRT

Since volume, gas constant R and T are constant, so we have

$\frac{p}{n]  = constant$

$\frac{44.7}{0.568} =\frac{22.8}{n}$

n = 0.289 mole

hence mass removed$\Delta M =( 0.568 - 0.289)*( molecular\ weight\ of \ nitrogen)$

$\Delta M = (0.568 - 0.289) *28 gm$

$\Delta M = 7.812 gm$