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Chemistry

Calculate the following :

The volume at s.t.p. occupied by a gas 'Q' originally occupying 153.7 cm3 at 287 K and 750 mm. pressure [vapour pressure of gas 'Q' at 287 K is 12 mm of Hg.]

Gas Laws

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Answer

Initial conditionsFinal conditions [S.t.p.]
P1 = Initial pressure of the gas = 750 mm - 12 mm = 738 mmP2 = Final pressure of the gas = 760 mm
V1 = Initial volume of the gas = 153.7 cm3V2 = Final volume of the gas = ?
T1 = Initial temperature of the gas = 287 KT2 = Final temperature of the gas = 273 K

By Gas Law:

P1×V1T1=P2×V2T2\dfrac{\text{P}1\times\text{V}1}{\text{T}1} = \dfrac{\text{P}2\times\text{V}2}{\text{T}2}

Substituting the values :

738×153.7287=760×V2273V2=738×153.7×273287×760V2=141.970 cm3\dfrac{738\times 153.7}{287} = \dfrac{760\times\text{\text{V}}2}{273} \\[1em] \text{V}2 =\dfrac{738\times 153.7 \times 273}{287\times 760} \\[1em] \text{V}_2 = 141.970 \text{ cm}^3

Therefore, Final volume of the gas = 141.970 cm3

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