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ICSE Class 9 Chemistry Question 26 of 29

Study of Gas Laws — Question 5

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5
Question

Question 1(1992)

A fixed volume of a gas occupies 760 cm3 at 27°C and 70 cm of Hg. What will be it's vol. at s.t.p.

Answer
Initial ConditionsS.T.P.
P1 = 70 cm of HgP2 = 76 cm of Hg
T1 = 27 + 273 = 300 KT2 = 273 K
V1 = 760 cm3V2 = ?

Using gas laws:

P1V1T1\dfrac{\text{P}_1{\text{V}_1}}{\text{T}_1} = P2V2T2\dfrac{\text{P}_2{\text{V}_2}}{\text{T}_2}

Substituting the values:

70×760300\dfrac{70 \times 760}{300} = 76×V2273\dfrac{76 \times {\text{V}_2}}{273}

Therefore:

V2=70×760×273300×76=637cm3\text{V}_2 =\dfrac{70 \times 760 \times 273}{300 \times 76} = 637 \text{cm}^3

Therefore, the volume occupied by the gas is 637 cm3.