A) Assuming constant temperature, as a helium balloon rises through the air: (Se

Question

A) Assuming constant temperature, as a helium balloon rises through the air: (Select all that apply.)
1.The air pressure outside the balloon decreases.
2.The volume of the balloon decreases.
3. The air pressure inside the balloon increases.
4.The air pressure inside the balloon decreases.
5.The air pressure outside the balloon increases.
6.The volume of the balloon increases.

B) An ideal gas at 26.7°C and a pressure 1.80 105 Pa is in a container having a volume of 1.00 L.
(a) Determine the number of moles of gas in the container. (Answer in mol)

(b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature. (answer in K)

C) Suppose the temperature of 4.51 L of ideal gas drops from 378 K to 275 K.
(a) If the volume remains constant and the initial pressure is atmospheric pressure, find the final pressure. (ANSWER IN Pa)

(b) Find the number of moles of gas. (ANSWER IN mol)

Explanation / Answer

A) 1. true 2. false 3. false 4. true 5. false 6. true B a) PV = nRT or you could use ratios each with one mole at STP as one of the parts 1/22.4*273/(273+26.7)*1.80/1.01 = 0.072 moles The pressure is making it more, the other two factors are making it less than a mole b) Pressure goes down by 1.01/1.8 this lowers t, volume doubles, this doubles t these are both in the numerator to the first power so 2*(1.01/1.8) = 1.222 Temp goes up by a factor of 1.222 (273+26.7)*1.222 = 336.3K C) a) PV=nRT only changing P and T. If T goes down by 275/378, P also goes down by same factor 101,325Pa*275/378 = 73,715 Pa b) (4.51/22.4)*(273/378) = 0.146 moles again I compare everything to one mole at stp, which occupies 22.4l

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