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

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

Answer 2

Answer:

23C

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: $8.31 \frac{Pa.m^{3} }{mol.K}$

we transform the liters to cubic meters:

$10 dm^3.\frac{1 m^3}{1000dm^3}= 0.01 m^3$

we clear the temperature and replace the data

$PV=nrT\\ T=\frac{PV}{nr}$

$T=\frac{123KPa.0.01 m^3}{8.31 \frac{Pa.m^3}{molK} . 0.5mol }= 296K$

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

$(296-273)= 23C$