Answer: The temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 2300 mm Hg = 3.02 atm (760mmHg=1atm)
V = Volume of gas = 15 L
n = number of moles = 0.6
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature = ?
[tex]T=\frac{PV}{nR}[/tex]
[tex]T=\frac{3.02atm\times 15L}{0.0821Latm/K mol\times 0.6mol}=920K[/tex]
Thus the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg is 920 K
