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

Answer:

$409.05K$

Explanation:

Hello,

The general law of the gases allows us to state:

$\frac{V_1P_1}{T_1}= \frac{V_2P_2}{T_2}$

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

$T_2=\frac{T_1P_2V_2}{P_1V_1}=\frac{303K*90kPa*3.75L}{100kPa*2.50L} =409.05K$

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