Question

T1 T2 Gay-Lussacs Law: Dont forget to convert C to K 1. 9.0L of a gas is exerts a pressure of 83.0 kPa at 35.0°C. What would be the required temperature (in ℃elsius) to change the pressure to standard pressure? 4
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Answer #1

According to this law,

Pressure is directly proportional to temperature

i.e, P1/T1= P2/T2

P1= 83.0 kpa = 0.819 atm

P2= 1 atm

T1= 35°C = 308 Kelvin

T2=?

T2= P2.T1/P1

=1* 308/0.819

= 376.0683 kelvin

i.e, temperature in degree celcious = kelvin temp - 273.15

= 102.9183°C

Required temperature is 102.9183° C

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