AP EAPCET 25-Aug-2021 Shift 2 Paper

Given, initial pressure, p1=p
Final pressure, pf=

p

2

Initial temperature T=400K
Final temperature T′=300K
Initial mass of gas, m=6g
From ideal gas equation, we get
pV=nRT
pV=

m

M

‌RT (∵

m

M

=n)
Initial condition, pV=

m

M

‌RT ...(i)
Final condition, p′V=

m′

M

RT′ ...(ii)
Dividing Eq. (ii) by Eq. (i), we get

p′

p

=

m′′T′

mT

⇒ m′=

p′

p

×

T

T′

×m=

p

2

×

1

p

×

400

300

×6
⇒ m′=4g
Mass of oxygen leaked, Δm=m−m′
⇒ Δm=6−4=2g