Answer:
320pka
Explanation:
This is a complete question;
If the temperature of a gas increased from 100 K to 200 K and the volume of a gas decreased from 20 L to 10 L, what is the new pressure if the original pressure was 80kPa?
EXPLANATION;
Gas laws can be regarded as aws which relate the pressure, volume, as well as temperature of a gas. It can be expressed as
P1V1/T1 = P2V2/T2
Where
V1= initial volume= 20L
V2 = final volume= 10L
T1 = initial temperature= 100K
T2 = final temperature= 200K
P2=final pressure=
P1= initial pressure= 80pka
Making P2 subject of the formula from the eqn, we have
P2= (P1*V1*T2=)/(T1*V2)
If we substitute the values we have
P2= (80 × 20×200)/(100×10)
P2= 320pka
Hence, the new pressure is 320pka
