Question

When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from ______ C to 87 C.

Answer 1

V1/T1=V2/T2
(15.5)/(360K)=(3.6)/(T2)
T2=83.61290K
T2=-189.3871 degrees Celsius

Answer 2

Hello!

When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from ______ C to 87 ºC.

We have the following data:

V1 (initial volume) = 3.6 L

V2 (final volume) = 15.5 L

T1 (initial temperature) = ? (answer in Celsius)

T2 (final temperature) = 87 ºC (to Kelvin)

TK = TC + 273.15 → TK = 87 + 273.15 → T2 (final temperature) = 360.15 K

We have an isobaric transformation, that is, when a certain mass under pressure maintains its constant pressure, on the other hand, as we increase the temperature, the volume increases and if we lower the temperature, the volume decreases and vice versa. We apply the data to the formula of isobaric transformation (Gay-Lussac), let us see:

`(V_1)/(T_1) =(V_2)/(T_2)`

`(3.6)/(T_1) =(15.5)/(360.15)`

multiply the means by the extremes

`15.5*T_1 = 3.6*360.15`

`15.5\:T_1 = 1296.54`

`T_1 = (1296.54)/(15.5)`

`T_1 \approx 83.647K`

Let's convert the Kelvin scales to Celsius, let's see:

`T\ºC = TK - 273.15`

`T\ºC = 83.647 - 273.15`

`T\ºC = - 189.503\to \boxed{\boxed{T_1\:(initial\:temperature) \approx\:-189.5\:\ºC}}\end{array}}\qquad\checkmark`

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I Hope this helps, greetings ... Dexteright02! =)