Question

10.0g of nitrogen dioxide gas has a volume of 6.3 L and is collected at 650 mm Hg what is the temperature

Answer 1

Step 1

Gas (dioxide gas - NO2) is assumed to be ideal. Therefore, it is applied:

p x V = n x R x T

p = pressure = 650 mmHg (p will be converted because R will be 0.082 atm x L/mol K)

So, 1 atm = 760 mmHg => 650 mmHg x (1 atm/760 mmHg) = 0.86 atm

V = volume = 6.3 L

n = number of moles = mass NO2/molar mass NO2 = 10.0 g/46 g/mol = 0.22 moles

(Molar mass NO2 = 46 g/mol)

R = gas constant = 0.082 atm x L/mol K

T = temperature = unknown

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Step 2

T is cleared:

$\begin{gathered} \frac{pxV}{nxR}=T \\ \frac{0.86\text{ atmx6.3 L}}{0.22\text{ moles x 0.082}\frac{atmxL}{molxK}}=T \\ 300.3\text{ K = T} \end{gathered}$

Answer: T = 300.3 K