A vessel has 6 g of oxygen at pressure P and temperature 400 K. A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is at P/2 and temperature 300 K?

For m grams of gas

\(PV = \frac{m}{M}RT\)…. (1)\\{P}’V = \frac{{m}’}{M}R{T}'\) … (2)

On dividing equation (2) by (1), we get,

\(\frac{{P}’}{P}=\frac{{m}’}{m}\frac{{T}’}{T}\)

On further calculation, we get,

\({m}’= \frac{{P}’}{P}\times \frac{T}{{T}’}\times m\)

On substituting the values, we get,

\({m}’= \left ( \frac{\frac{P}{2}}{P} \right )\times \frac{400}{300}\times 6\)

= 4g

Mass of oxygen leaked:

\(\Delta m=m-{m}’=6-4=2g\)

Hence, mass of oxygen leaked is 2g 

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question