Question

At what temperature will 6.21 g of oxygen gas exert a pressure of 5.00 atm in a 10.0-L container?

Answer 1

Answer:The value of the temperature is 2,870.072°C.

Step-by-step explanation:

Moles of oxygen = n =
`(6.21 g)/(32 g/mol)=0.1940 mol`

Pressure of the gas = 5.00 atm

Volume of the gas = 10.0 L

Temperature of the gas = T

`PV=nRT`

`5.00 atm* 10.0L=0.1940 mol* 0.0820 atm L/mol K* T`

`T=(5.00 atm* 10.0L)/(0.1940 mol* 0.0820 atm L/mol K)=3,143.072 K=2,870.072^oC`

(T(K)=T(°C)-273)

The value of the temperature is 2,870.072°C.

Answer 2

PV = nRT
P = 5 atm
V = 10 L

molecular mass of oxygen gas = 16 X 2 =32 g/mole

so number of moles of oxygen gas =

6.21/32 = 0.194
so n = 0.194
R = 0.0821 L atm K^-1 mole^-1
T = ? K



5 X 10 = 0.194 X 0.0821 X T

50 = 0.0159 X T

T = 50/0.0159 = 3144.654 K

in degree c = 3144.654 - 273
= 2871.65 degree c