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ICSE Class 9 Chemistry Question 23 of 29

Study of Gas Laws — Question 6

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Question 6

To what temperature must a gas at 300 K be cooled down in order to reduce its volume to 13\dfrac{1}{3}rd of its original volume, pressure remaining constant?

Answer

V1 = Initial volume of the gas = V

T1 = Initial temperature of the gas = 300 K

V2 = Final volume of the gas = 13\dfrac{1}{3}V

T2 = Final temperature of the gas = ?

By Charles' Law:

V1T1=V2T2\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}

Substituting the values :

V300=13VT2T2=3003T2=100 K\dfrac{\text{V}}{300} = \dfrac{\dfrac{1}{3}\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{300}{3} \\[1em] \text{T}_2 = 100 \text{ K}

∴ Final temperature = 100 K