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Question

To what temperature must chlorine gas at 132°C be colled in order to reduce its volume to one fifth of ots original volume

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Solution

Dear student
given that
t₁ = 132 oC​​ = (132 + 273) K = 405 K
let v₁ = V
∴ v₂ = V/5,
t₂ = The temperature at which gas is cooled down
​​​​​according to Charles' Law
v₁/t₁ = v₂/t₂ (at Constant pressure)
subsituting values,
V/405 = V/5/t₂
V/300 = V/5t₂
5t₂ = 405
∴ t₂ = 81 K = (81 - 273) oC = -192 oC
the chlorine gas should be cooled to -192 oC such that volume becomes 1/5th

Regards

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