A gas has a volume of 5.20L at 0•C. What final temperature in degrees celcius is

Question

A gas has a volume of 5.20L at 0•C. What final temperature in degrees celcius is needed to cause the volume of the gas to change to the following, if n (number of moles of gas) and P are not changed?
* express all answers in 2 sig fig*
A.) 12L T1= ? Degrees Celcius
B.) 1300 mL T2= ? Degrees Celcius
C.) 20L T3= ? Degrees Celcius
D.) 42.0 mL T4= ? Degrees Celcius A gas has a volume of 5.20L at 0•C. What final temperature in degrees celcius is needed to cause the volume of the gas to change to the following, if n (number of moles of gas) and P are not changed?
* express all answers in 2 sig fig*
A.) 12L T1= ? Degrees Celcius
B.) 1300 mL T2= ? Degrees Celcius
C.) 20L T3= ? Degrees Celcius
D.) 42.0 mL T4= ? Degrees Celcius
* express all answers in 2 sig fig*
A.) 12L T1= ? Degrees Celcius
B.) 1300 mL T2= ? Degrees Celcius
C.) 20L T3= ? Degrees Celcius
D.) 42.0 mL T4= ? Degrees Celcius

Explanation / Answer

V = 5.2

T = 0C or 273K

Tf = ?

a)

V = 12L

V1/T1 = V2/T2

5.2L/273K = 12L/T2

T2 = (12/5.2)*273 = 630K = 630-273 = 357°C

B)

V1/T1 = V2/T2

5.2ml/273K = 1.3L/T2

T2 = (1.3/5.2)*273 = 68.25 K = -204.75 C

C)

V1/T1 = V2/T2

5.2ml/273K = 20L/T2

T2 = (20/5.2)*273 = 1050 K  = -777°C

D)

V1/T1 = V2/T2

5.2ml/273K = 0.042L/T2

T2 = (0.042/5.2)*273 = 2.205 K = -270.795°C

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