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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?

Solutions

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 2 years ago
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