Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.