Question

Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0L. What temperature must be maintained on 0.500 miles of gas in order to maintain this pressure? Express the temperature in degrees Celsius?

Answer 1

23C

Step-by-step explanation:

As oxygen is in a gaseous state, and with the data that gives us the problem we must use the equation of ideal gases

`PV=nrT`

P= pression

V= volumen

n= moles

r= constant of ideal gases

T= temperature

data:

Pression: 123 KPa

Volumen: 10.0L

moles: 0.5 moles

constant of ideal gases:
`8.31 (Pa.m^(3) )/(mol.K)`

we transform the liters to cubic meters:

`10 dm^3.(1 m^3)/(1000dm^3)= 0.01 m^3`

we clear the temperature and replace the data

`PV=nrT\\ T=(PV)/(nr)`

`T=(123KPa.0.01 m^3)/(8.31 (Pa.m^3)/(molK) . 0.5mol )= 296K`

The temperature is 296 K but we have to expressed in Celsius

`(296-273)= 23C`

Answer 2

Given:

P = 123 kPa
V = 10.0 L
n = 0.500 moles
T = ?

Assume that the gas ideally, thus, we can use the ideal gas equation:

PV = nRT

where R = 0.0821 L atm/mol K

123 kPa * 1 atm/101.325 kPa * 10.0 L = 0.500 moles * 0.0821 Latm/molK * T

solve for T

T = 295.72 K