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NikAS [45]
3 years ago
12

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 t

o 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?
Chemistry
1 answer:
leonid [27]3 years ago
6 0

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.

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