At what temperature will 6.21 g of oxygen gas exert a pressure of 5.00 atm in a 10.0-L container?
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
Answer:The value of the temperature is 2,870.072°C.
Explanation:
Moles of oxygen = n =\(\frac{6.21 g}{32 g/mol}=0.1940 mol\)
Pressure of the gas = 5.00 atm
Volume of the gas = 10.0 L
Temperature of the gas = T
\(PV=nRT\)
\(5.00 atm\times 10.0L=0.1940 mol\times 0.0820 atm L/mol K\times T\)
\(T=\frac{5.00 atm\times 10.0L}{0.1940 mol\times 0.0820 atm L/mol K}=3,143.072 K=2,870.072^oC\)
(T(K)=T(°C)-273)
The value of the temperature is 2,870.072°C.