The temperature of the gas sample is 813 K.
Explanation:
We have to use the ideal gas equation to find the temperature of the gas sample.
The ideal gas equation is PV = nRT
Pressure, P = 429 mm Hg = 0.56 atm
Volume, V = 560 mL = 0.56 L
R = gas constant = 0.08205 L atm mol⁻¹K⁻¹
Mass = 0.211 g
Molar mass of carbon di oxide = 44.01 g / mol
Moles, n = [tex]$\frac{given mass}{molar mass} = \frac{0.211 g}{44.01 g/mol}[/tex]
= 0.0047 mol
Now, we have to plugin the above values in the above equation, we will get the temperature as,
[tex]$T= \frac{PV}{nR}[/tex]
T = [tex]$\frac{0.56 \times 0.56}{0.08205 \times 0.0047}[/tex]
= 813 K
So the temperature of the gas sample is 813 K.
