Question

A gas is contained in a thick-walled balloon. When the pressure change from 100 kPa to 90.0 kPa, the volume changes from 2.50 L to 3.75 L and the temperature changes from 303 K to _____ K.

Answer 1

`409.05K`

Step-by-step explanation:

Hello,

The general law of the gases allows us to state:

`(V_1P_1)/(T_1)= (V_2P_2)/(T_2)`

Now, as the temperature at the second state is the unknown, we solve for it and get:

`T_2=(T_1P_2V_2)/(P_1V_1)=(303K*90kPa*3.75L)/(100kPa*2.50L) =409.05K`

Best regards.

Answer 2

The temperature change is calculated using the combined gas law
that is P1V1/T1 =P2V2/T2
P1= 100KPa
P2=90kpa
v1= 2.50 L
v2= 3.75 L
T1= 303 K
T2=?

T2 is therefore = P2V2T1/P1V1
=( 90 x 3.75 x303)/ (100 x2.50) = 409.05 K