Ask a New Question

Question

A 813. mL sample of argon gas at 452. K is cooled at constant pressure until the volume becomes 356. mL. What is the new gas temperature in Kelvin?
10 years ago

Answers

Pre-med student
PV=PV
.813L*452=.356X
10 years ago
DrBob222
There is no P listed and the formula is not PV = PV. However the answer is correct.
It is (V1/T1) = (V2/T2)
(813/452) = (356/T2)
T2 = 813*452/356
10 years ago

Related Questions

A sample of argon at 300.°C and 50.0 atm pressure is cooled in the same container to a temperature o... A sample of argon gas has a volume of 715 mL at a pressure of 1.34 atm and a temperature of 144 Cels... A 1.30-L sample of argon gas is at 1.02 atm and 21.5 degrees Celsius. a. what mass of argon is in th... A 15.0L sample of argon has a pressure of 28.0 atm . What volume would this gas occupy at 9.90 atm?... An 0.727 mol sample of argon gas at a temperature of 22.0 °C is found to occupy a volume of 26.2 lit... A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temp of 112 degrees Cel... A sample of argon at 2.0 atm pressure is heated until its temperature is raised from 60 °C to 90 °C.... A sample of argon gas has a volume of 735 mL m L at a pressure of 1.20 atm a t m and a temp... A sample of argon-39 had an original mass of 1578 grams. After 538 years, the sample is 394.5 grams.... A sample of argon-39 had an original mass of 1578 grams. After 538 years, the sample is 394.5 grams....
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use