A given mass of gas occupies 919.0ml in dry state at N.T.P. The same mass when collected over water at 15°c and 0.987 bar pressure occupies one litre volume. Calculate the vapour pressure of water at 15°c
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Explanation:
Let us consider the
P
drygas
=P
total
=P
watervapour
Given that:
(At STP)
V
1
=919mL
V
2
=100mL
P
1
=760mmHg
P
2
=?(drygas)
T
1
=273K
T
2
=273+15=288K
P
total
=750mmHg
Using gas equation
T
1
p
1
v
1
=
T
2
p
2
v
2
implies that
273
760×919
=
288
p
2
×1000
implies that
p
2
=736.82mmHg
Hence
P
drygas
=P
total
−P
watervapour
implies that
P
watervapour
=750−736.82
13.18mm
HOPE IT HELP YOU!
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