TS EAMCET 04 May 2019 Shift 2 Solved Paper

When ∆V volume of gas is let out, the pressure of gas is
‌ decreased to ‌∆p.‌ So, ‌
V2=V−∆V
And,
p2=p−∆p
Also,
T1=T2
By an ideal gas equation
‌

p1V1

T1

‌‌=‌

p2V2

T2

‌

p1

p2

‌‌=‌

V2

V1

‌

p

p−∆p

‌‌=‌

V−∆V

V

‌

p

p−∆p

‌‌=1−‌

∆V

V

Simplify further,
‌

∆V

V

=1−‌

p

p−∆p

‌

∆V

V

=‌

p−∆p−p

p−∆p

∆V=‌

∆pV

p−∆p

Therefore, the mass of the released gas is,
mRG‌‌=ρ∆V
‌‌=ρ‌

∆pV

p−∆p

‌‌≈‌

ρ∆pV

p