If a contained ideal gas decreases the volume by a factor of twoand then triples...

If a contained ideal gas decreases the volume by a factor of two and then triples the temperature, what happens to the pressure?

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Expert Solution

let P1 is the initial pressure
V1 is the initial volume
T1 is the initial temeperature

final volume, V2 = V1/2

final temperature, T2 = 3*T1

P2 = ?

use, idela gas equation, P*V = n*R*T

P*V/T = n*R

P*V/T = constant

so, P2*V2/T2 = P1*V1/T1

P2 = P1*(V1/V2)*(T2/T1)

= P1*(V1/(V1/2))*(3*T1/T1)

= P1*2*3

= 6*P1

so, pressure becomes 6 times

genius_generous answered 1 year ago

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