Question

A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated to 100 oC with the stopcock open to the atmosphere (like your Charles Law lab). The stopcock is then closed and the flask cooled to its original temperature (20 oC). K=1.32x10-23 J/k. (a) What is the final pressure of the oxygen in the flask (with the stopcock closed)? (b) How many grams of oxygen remain in the flask?

Answer 1

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = $P_1=1.0 atm$

Initial temperature of the oxygen gas = $T_1=20^oC =293.15 K$

Final pressure of the oxygen gas = $P_2=?$

Final temperature of the oxygen gas = $T_2=100^oC =373.15 K$

Using Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm$

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

$P_1V_1=nRT_1$ (ideal gas equation)

$n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol$

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.