Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Charle's Law states that the volume varies directly with it's temperature and can be expressed as :
[tex]V=kT[/tex]Where V is the volume, T is the temperature and k is some constant
From the given problem, we have :
V1 = 0.8 liters
T1 = 400 degrees
V2 = 0.3 liters
Since k is a constant,
[tex]V_1=kT_1[/tex][tex]V_2=kT_2[/tex]We can express both equation as k = V/T
[tex]k=\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]Substitute the given values to the formula :
[tex]\frac{0.8}{400}=\frac{0.3}{T_2}[/tex]Then solve for the value of T2 :
Simplify the equation by multiplying 400T2 to both sides of the equation :
[tex](400T_2)\times\frac{0.8}{400}=(400T_2)\times\frac{0.3}{T_2}[/tex][tex]0.8T_2=120[/tex][tex]T_2=\frac{120}{0.8}=150[/tex]Therefore the answer is 150 degrees
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.