Question

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

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

`(3.75)/(100) =(6.52)/(T_2)`

`3.75*T_2 = 100*6.52`

`3.75\:T_2 = 652`

`T_2 = (652)/(3.75)`

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

_________________________

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

Answer 2

The temperature will change from 100K to 173.87 K

calculation

by use of law that is V1/T1=V2/T2

V1=3.75 L

T1=100k

V2=6.53 L

T2=?

make T2 the subject of the formula

T2=(V2 xT1)V1

=6.52 x100/3.75=173.87K