Question

A vessel has 6g of oxygen at a pressure p and temperature 400 k

Answer 1

The missing part of the question is given below.

"A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is P/2 and temperature is 300K."

Answer : 2 grams of oxygen leaks out of the vessel.

Explanation :

Here the volume of the vessel is kept constant.

Step 1: Find the volume of the vessel.

We have been given ,

Pressure = P

Temperature = 400 K

Let us find moles of oxygen.

Mole is calculated using following formula.

$mole = \frac{grams}{MolarMass}$

Molar mass of oxygen gas is 32 /mol

$mole = \frac{6g}{32g/mol}$

$mole = 0.188$

So we have n = 0.188 mol.

The ideal gas equation is written as

$PV = nRT$

Let us modify the equation to solve for V

$V = \frac{nRT}{P}$

Let us plug in the given values.

$V = \frac{0.188 \times 0.0821 \times 400}{P}$

On solving further we get,

$V = \frac{6.174}{P} ................ (equation 1)$

Let us call this equation 1.

Step 2 : Set up a new equation for V after the gas leaks out.

When the hole is made in the vessel, some of the gas leaks out .

Due to this pressure changes to $pressure = \frac{P}{2}$

and temperature changes to $T = 300 K$

Let us assume n moles of gas are left.

The new equation for V can be written as

$V = \frac{n \times 0.0821 \times 300}{P/2}$

On further simplifying we get,

$V = \frac{n \times 0.0821 \times 300 \times 2 }{P}$

$V = \frac{n \times 49.26}{P} .............(equation 2)$

Let us call this equation 2.

Step 3 : Equate the volumes in equation 1 and 2.

We know that the volume of the vessel remains constant. Therefore we can equate equation 1 and equation 2.

$\frac{6.174}{P} = \frac{n \times 49.26}{P}$

P can be cancelled out.

$6.174 = n \times 49.26$

n = 0.125 mol

Amount of gas left in the vessel is 0.125 moles.

The initial amount of gas was 0.188 moles.

Amount of gas left = 0.188 - 0.125 = 0.063 moles.

Let us convert this to grams of oxygen

$0.063 mol \times \frac{32g}{mol} = 2.0 g$

Therefore we can say, 2 grams of oxygen leaks out.

Answer 2

$From\quad the\quad given,\\ Mass\quad of\quad oxygen\quad =\quad 6g\\ Inittial\quad temperature\quad =\quad 500\quad K\\ Final\quad temperature\quad \quad -\quad 300\quad K\\ PV\quad =\quad \frac { m }{ M } RT\\ m\quad =\quad 6tg\\ Intially\quad ,\quad PV\quad =\quad \frac { 6 }{ M } R\quad \times \quad 500\quad ...............................(i)$

$Finaly,\quad \frac { P }{ 2 } V\quad =\quad \frac { (6-x) }{ M } R\quad \times \quad 300\\ (if\quad "x"\quad gas\quad leaks\quad out)\\ Hence,\quad \\ 2\quad =\quad \frac { 6 }{ 6-x } \times \frac { 5 }{ 3 } \\ x\quad =\quad 1\quad g\\ Therefore,\quad leaked\quad amount\quad of\quad oxygen\quad =\quad 6g$