Question 12.26: Applying the Ideal Gas Law to a Mixture of Gases A gaseous m......
Applying the Ideal Gas Law to a Mixture of Gases
A gaseous mixture consists of 3.00 g of N _2 and 7.00 g of Ne. What is the volume, in liters, occupied by this mixture if it is under a pressure of 2.00 atm and at a temperature of 27°C?
We first calculate the number of moles of each of the gases present and then add these quantities together to obtain the total moles of gas present.
n_{\textrm{N}_2} : 3.00 \cancel{\textrm{g N}_2} \times \frac{1 \textrm{mole N}_2}{28.02 \cancel{\textrm{g N}_2}} = 0.10706638 mole N _2 (calculator answer)
= 0.107 mole N _2 (correct answer)
n_{\textrm{Ne}} : 7.00 \cancel{\textrm{g Ne}} \times \frac{1 \textrm{mole Ne}}{20.18 \cancel{\textrm{g Ne}}} = 0.34687809 mole Ne (calculator answer)
= 0.347 mole Ne (correct answer)
n_{\textrm{total}} = n_{\textrm{N}_2} + n_{\textrm{Ne}} = (0.107 + 0.347) mole = 0.454 mole (calculator and correct answer)
Rearranging the ideal gas law to isolate V on the left side of the equation gives
V = \frac{nRT}{P}All of the quantities on the right side of the equation are known.
n_{\textrm{total}} = 0.454 mole, R = 0.08206 \frac{\textrm{atm} \cdot \textrm{L}}{\textrm{mole} \cdot \textrm{K}} T = 300 K, P = 2.00 atm
Substituting in our equation, cancelling units, and doing the arithmetic gives
V = \frac{(0.454 \cancel{\textrm{mole}}) \Bigl(0.08206 \frac{\cancel{\textrm{atm}} \cdot \textrm{L}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (300 \cancel{\textrm{K}})}{2.00 \cancel{\textrm{atm}}}= 5.588286 L (calculator answer)
= 5.59 L (correct answer)
Answer Double Check: Is the magnitude of the numerical answer reasonable? Yes. Approximately 0.5 total moles of gas are present, which would give 11.2 L of gas at STP conditions. The conditions differ from STP mainly in pressure, which is double that at STP. This would decrease the volume by a factor of 2 to 5.6 L. The calculated answer is consistent with this rough analysis.