Asked by Mayank Goswami
the volume at s.t.p. occupied by a gas q originally occupying 153.7 cc at 287 k and 750mm pressure, vapour pressure of gas q at 287k is 12 mm of hg
Answers
Answered by
DrBob222
You didn't ask a question.
Answered by
hihahahaa
Explanation:
Given data:
Volume of gas = 153.7 cm^3
Temperature "T" = 287 K
Pressure "P" = 750 mm
P1V1/ T1 = P2V2 / T2
We have p1 = 750 + vapour pressure of gas at 287 K
= (750 + 12) mm of Hg = 762 mm of Hg
V1 = 153.7 cm^3 and T1 = 287 K
p2 = 760 mm of Hg , T2 = 273 K
(As STP conditions are 1 atm pressure which is equal to 760 mm of Hg and 273K temperature)
And V2 = ?
Therefore V2 = (p1V1) / T1 X (T2 / p2)
= (762 X 153.7 X 273) / ( 287 X 760)
= 146.6 cm^3
Hence the second Volume is 146.6 cm^3.
Given data:
Volume of gas = 153.7 cm^3
Temperature "T" = 287 K
Pressure "P" = 750 mm
P1V1/ T1 = P2V2 / T2
We have p1 = 750 + vapour pressure of gas at 287 K
= (750 + 12) mm of Hg = 762 mm of Hg
V1 = 153.7 cm^3 and T1 = 287 K
p2 = 760 mm of Hg , T2 = 273 K
(As STP conditions are 1 atm pressure which is equal to 760 mm of Hg and 273K temperature)
And V2 = ?
Therefore V2 = (p1V1) / T1 X (T2 / p2)
= (762 X 153.7 X 273) / ( 287 X 760)
= 146.6 cm^3
Hence the second Volume is 146.6 cm^3.
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