Chemistry - Ideal Gas Question with Solution | TestHub
Paragraph for Questions 10 and 11
A 4.92 L Duma's bulb contains nitrogen gas at 1.0 atm. On adding 0.05 mole of gas it is necessary to cool the bulb to a temperature of 27° to maintain the same pressure. ( = 0.082 L atm/K mol)
The bulb at 27° (with the added) is heated back to its initial temperature at constant volume, and then 0.10 mol of gas is vented at that temperature. The final pressure in the bulb is
Options:
Answer:
Solution:
Initially / = 1.0 x 4.92/0.082 = 60 mol K. After adding 0.05 mol , ( + 0.05) x 300 = 60, so = 0.15 mol and = 400 K.
After adding : 0.20 mol at 27° and 1.0 atm in 4.92 L.
Heated back to 400 K at constant volume, the gas is still 0.20 mol. Venting 0.10 mol leaves 0.10 mol at 400 K in 4.92 L.
= / = 0.10 x 0.082 x 400/4.92 = 0.667 atm.
The values 1.00 and 1.33 atm come from not venting or using the wrong amount.
Answer: ().
