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ICSE Class 9 Chemistry Question 17 of 29

Study of Gas Laws — Question 26

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Question

Question 26

(a) Calculate the volume of S.T.P. of 380 cm3 of carbon dioxide at 91°C and 72 mm of Hg pressure.

(b) What will be its volume at 273 K and 70 cm of Hg pressure ?

Answer

(a)

Initial conditionsFinal conditions
P1 = Initial pressure of the gas = 72 mm HgP2 = Final pressure of the gas = 760 mm Hg
V1 = Initial volume of the gas = 380 cm3V2 = Final volume of the gas = ?
T1 = Initial temperature of the gas = 91°C = 91 + 273 = 364 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 :

72×380364=760×V2273V2=72×380×273364×760V2=27 cm3\dfrac{72 \times 380}{364} = \dfrac{760 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{72 \times 380 \times 273}{364\times 760} \\[1em] \text{V}_2 = 27 \text{ cm}^3 \\[1em]

(b)

Initial conditionsFinal conditions
P1 = Initial pressure of the gas = 72 mm of HgP2 = Final pressure of the gas = 70 cm Hg = 700 mm of Hg
V1 = Initial volume of the gas = 380 cm3V2 = Final volume of the gas = ?
T1 = Initial temperature of the gas = 91°C = 91 + 273 = 364 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 :

72×380364=700×V2273V2=72×380×273364×700V2=29.314 cm3\dfrac{72 \times 380}{364} = \dfrac{700 \times \text{V}_2}{273} \\[1em] \text{V}_2 = \dfrac{72 \times 380 \times 273}{364\times 700} \\[1em] \text{V}_2 = 29.314 \text{ cm}^3 \\[1em]