BSc Nursing GNM 2019 Solved Paper

For 1g mole of gas, gas equation is
pV=RT
For 1g of a gas, gas equation becomes
pV=rT
For m grams of a gas, the equation is
pV=mrT
Such that, ‌‌m=‌

pV

rT

Since, bulbs are in equal volume and they arefilled under identical conditions at NTP.
∴‌‌p=76cm‌ of ‌Hg,
T‌‌=0∘C=273K
If V is the volume of each bulb, then
m=‌

76V

r×273

∴ Total mass of gas in two bulbs is
2m=‌

2×76V

r×273

. . . (i)
When one bulb is immersed in ice, the gas will contract and when other is immersed in boiling water, the gas will expand and will flow from bulb in boiling water to bulb in ice, until the pressure in both the bulbs becomes same. Let p be the final pressure in the two bulbs.
∴ Mass of the bulb in boiling water.
m1=‌

pV

r×(273+100)

=‌

pV

r×373

...‌ (ii) ‌
and mass of the bulb in ice,
m2=‌

pV

r×(273+0)

=‌

pV

r×273

...‌ (iii) ‌
Total mass of the gas in two bulbs,
2m=m1+m2
On substituting values of 2m,m1 and m2
from Eqs. (i), (ii) and (iii) in above equation, we get
‌

2×76V

r×273

=‌

pV

r×373

+‌

pV

r×273

‌ or ‌‌‌‌

152

273

=‌

p

373

+‌

p

273

‌ or ‌‌‌‌

152

273

=p(‌

1

373

+‌

p

273

)
or ‌‌‌

152

273

=‌

p×646

373×273

or ‌‌p=‌

152×373×273

273×646

∴‌‌p=87.76cm of Hg