Question

A vessel has 6 gram Oxygen gas at pressure P and temperature 400K. A hole is made in the vessel and Oxygen gas leaked find out how much amount of Oxygen gas leaked out if final pressure is P/2 and temperature is 300K.​

Answer 1

Answer:

2g

Explanation:

Given

$P_{1} = P \\ P_{2} = \frac{P}{2} \\ V_{1} = V \\ V_{2} = V \\ T_{1} = 400 \: K \\ T_{2} = 300 \: K$

$\bold{Now, \: we \: need \: n_{1} \: and \: n_{2}}$

$n_{1} = \frac{6}{32} \\ n_{2} = \frac{M}{32}$

We know that

$\frac{P_{1}V_{1}}{n_{1}T_{1}} = \frac{P_{2}V_{2}}{n_{2}T_{2}}$

On putting all the values

$= > \frac{P \times V}{400 \times \frac{6}{32} } = \frac{ \frac{P}{2} \times V}{300 \times \frac{M}{32} }$

$= > \frac{ \cancel{P} \times \cancel{V} \times \cancel{32}}{ 400 \times 6} = \frac{ \cancel{P} \times \cancel{V} \times \cancel{32}}{ 2 \times {300} \times M} \\ = > 2400 = 600M \\ = > M = \frac{ \cancel{2400}}{ \cancel{600}} \\ = > \boxed{M = 4g}$

$\bold{Amount \: of \: leaked \: gas = 6g - 4g} \\ \bold{Amount \: of \: leaked \: gas = 2g}$

Answer 2

• Let Initial Pressure be P1 and Final Pressure be P2.

Also,

• P1 = P

• P2 = P/2

• Initial Temperature = 400K

• Final Temperature = 300K

We know that..

》 PV = nRT

=> V = $\dfrac{nRT}{P}$

Here V is volume which is constant for both.

So,

=> V1 = $\dfrac{n1R1T1}{P1}$ ______ (eq 1)

Similarly,

=> V2 = $\dfrac{n2R2T2}{P2}$ ______ (eq 2)

From both equations we have,

=> $\dfrac{n1R1T1}{P1}$ = $\dfrac{n2R2T2}{P2}$

=> $\dfrac{n1}{n2}$ = $\dfrac{P1T2}{P2T1}$

(As R is constant so cancel out)

=> $\dfrac{n1}{n2}$ = $\dfrac{P(300)}{P/2(400)}$

=> $\dfrac{n1}{n2}$ = $\dfrac{3}{2}$

Also, we know that.. here 'n' is number of moles.

Number of moles = Mass/Molar Mass

Mass = m

Molar Mass = m.m

=> $\dfrac{n1}{n2}$ = $\dfrac{m1/(m.m1)}{m2/(m.m2)}$

=> $\dfrac{3}{2}$ = $\dfrac{m1}{m2}$

=> $\dfrac{3}{2}$ = $\dfrac{6}{m2}$

(m1 = 6 g GIVEN)

=> m2 = 4 g

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Remaining gas = 4 g and Leaked gas = 6 - 4 = 2 g

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