Question

when the volume of a gas is changed from 3.75 L to 6.52 L the temperature will change from 100k to blank k

Answer 1

Hello!

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 have the following data:

V1 (initial volume) = 3.75 L

V2 (final volume) = 6.52 L

T1 (initial temperature) = 100 K

T2 (final temperature) =? (in Kelvin)

We apply the data to the formula of isobaric transformation (Gay-Lussac), let us see:

$\dfrac{V_1}{T_1} =\dfrac{V_2}{T_2}$

$\dfrac{3.75}{100} =\dfrac{6.52}{T_2}$

$3.75*T_2 = 100*6.52$

$3.75\:T_2 = 652$

$T_2 = \dfrac{652}{3.75}$

$\boxed{\boxed{T_2 \approx 173.87\:K}}\Longleftarrow(final\:temperature)\end{array}}\qquad\checkmark$

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

Answer 2

Guy-Lussac's Law states that the volume and the temperature are directly proportional given that the pressure remains constant.

For this problem, we will assume constant pressure. Based on the law:
(Volume/Temperatur)1 = (Volume/Temperature)2
(3.75/100) = (6.52/T)
T = 166.667 kelvin