# Temperature and Degrees Celsius

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

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