Answer: [tex]15.5^0C[/tex]
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 0.987 atm
V = Volume of gas = 12 L
n = number of moles = 0.50
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature = ?
[tex]T=\frac{PV}{nR}[/tex]
[tex]T=\frac{0.987atm\times 12L}{0.0820 L atm/K mol\times 0.50mol}=288.5K=(288.5-273)^0C=15.5^0C[/tex]
Thus the temperature of a 0.50 mol sample of a gas at 0.987 atm and a volume of 12 L is [tex]15.5^0C[/tex]
