Question

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

Answer 1

Answer:

T1 = 300 K

T2 = ?

__ __ __ __ __ __ __

V1 = V (let)

V2 = 1/3 of V = V/3

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Applying Charle's law of constant pressure,

$\frac{v1}{t1} \: = \: \frac{v2}{t2} \\ \\ = > \frac{v}{300} = \frac{ \frac{v}{3} }{t2} \\ \\ = > t2 = \frac{v}{3} \times \frac{300}{v} \\ = > \: \: t2 \: = \: 100 \: \: k$

Answer 2

$hello$

Answer : 100K

Explanation : V × T

= V¹/V² = T¹/ T²

= V/V/3 = 300/T²

= T² = 100K

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