Temperature and Degrees Celsius
Megan Potter
Joule’s Jungle Honors Extension
1. A gas occupies 0.60 m3 at a 5.0 atm. If the temperature of the gas remains the same and the pressure decreases to 2.5 atm, what would be the new volume occupied by the gas?
V1=.60 m3
P1=5.0 atm
P2= 2.5 atm
V2= ?
P1 V1=P2 V2
V2= (P1 V1)/ (P2)
V2= (5.0 atm)(.60 m3)/ (2.5 atm)
V2= 1.2 m3
2. Given a volume of 1000. cm3 of an ideal gas at 300. K, what volume would it occupy at a temperature of 600. K?
V1= 1000 cm3
T1= 300 K
T2= 600 K
V2= ?
V1/T1= V2/ T2
V2= (V1/ T1) (T2)
V2= (1000 cm3/ 300 K) (600 K)
V2= 2000 cm3
3. An ideal gas occupies a volume of 0.60 m3 at 5.0 atm and 400. K. What volume does it occupy at 4.0 atm and a temperature of 200. K?
V1=.60 m3
P1=5.0 atm
T1= 400 K
P2= 4.0 atm
T2= 200 K
V2= ?
P1 V1/T1 =P2 V2/T1
V2= (P1V1/T1)(T2)/ (P2)
V2= ((5.0 atm * .60 m3)/ 400 K) (200 K) / (4.0 atm)
V2= .375 m3
4. If a 10. m3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken to a high altitude where the pressure is 400. mm Hg and a temperature of -23 degrees Celsius, what volume will it occupy? (Hint: remember the temperature must be in Kelvin)
V1=10 m3
P1=760 mm
T1= 27 degrees Celsius or 300 K
P2= 400 mm
T2= -23 degrees Celsius or 250 K
V2= ?
P1 V1/T1 =P2 V2/T1
V2= (P1V1/T1)(T2)/ (P2)
V2= ((760 mm * 10m3)/ 300 K) (250 K) / (400 mm)
V2= 15.83 m3
Joule’s Jungle Honors Extension
1. A gas occupies 0.60 m3 at a 5.0 atm. If the temperature of the gas remains the same and the pressure decreases to 2.5 atm, what would be the new volume occupied by the gas?
V1=.60 m3
P1=5.0 atm
P2= 2.5 atm
V2= ?
P1 V1=P2 V2
V2= (P1 V1)/ (P2)
V2= (5.0 atm)(.60 m3)/ (2.5 atm)
V2= 1.2 m3
2. Given a volume of 1000. cm3 of an ideal gas at 300. K, what volume would it occupy at a temperature of 600. K?
V1= 1000 cm3
T1= 300 K
T2= 600 K
V2= ?
V1/T1= V2/ T2
V2= (V1/ T1) (T2)
V2= (1000 cm3/ 300 K) (600 K)
V2= 2000 cm3
3. An ideal gas occupies a volume of 0.60 m3 at 5.0 atm and 400. K. What volume does it occupy at 4.0 atm and a temperature of 200. K?
V1=.60 m3
P1=5.0 atm
T1= 400 K
P2= 4.0 atm
T2= 200 K
V2= ?
P1 V1/T1 =P2 V2/T1
V2= (P1V1/T1)(T2)/ (P2)
V2= ((5.0 atm * .60 m3)/ 400 K) (200 K) / (4.0 atm)
V2= .375 m3
4. If a 10. m3 volume of air (acting as an ideal gas) is at a pressure of 760 mm and a temperature of 27 degrees Celsius is taken to a high altitude where the pressure is 400. mm Hg and a temperature of -23 degrees Celsius, what volume will it occupy? (Hint: remember the temperature must be in Kelvin)
V1=10 m3
P1=760 mm
T1= 27 degrees Celsius or 300 K
P2= 400 mm
T2= -23 degrees Celsius or 250 K
V2= ?
P1 V1/T1 =P2 V2/T1
V2= (P1V1/T1)(T2)/ (P2)
V2= ((760 mm * 10m3)/ 300 K) (250 K) / (400 mm)
V2= 15.83 m3