a A balloon whose volume is 800 m3 is to be filled with hydrogen at atmospheric

Question

a

A balloon whose volume is 800 m3 is to be filled with hydrogen at atmospheric pressure (1.01×105Pa).

Part A

If the hydrogen is stored in cylinders with volumes of 1.95 m3 at a gauge pressure of 1.23×106 Pa , how many cylinders are required? Assume that the temperature of the hydrogen remains constant.

31.153

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Part B

What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at 15.0 C?The molar mass of hydrogen (H2) is 2.02 g/mol. The density of air at 15.0 C and atmospheric pressure is 1.23 kg/m3.

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Part C

What weight could be supported if the balloon were filled with helium (with a molar mass of 4.00 g/mol) instead of hydrogen, again at 15.0 C?

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A balloon whose volume is 800 m3 is to be filled with hydrogen at atmospheric pressure (1.01×105Pa).

Part A

If the hydrogen is stored in cylinders with volumes of 1.95 m3 at a gauge pressure of 1.23×106 Pa , how many cylinders are required? Assume that the temperature of the hydrogen remains constant.

N =

31.153

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Part B

What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at 15.0 C?The molar mass of hydrogen (H2) is 2.02 g/mol. The density of air at 15.0 C and atmospheric pressure is 1.23 kg/m3.

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Part C

What weight could be supported if the balloon were filled with helium (with a molar mass of 4.00 g/mol) instead of hydrogen, again at 15.0 C?

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Explanation / Answer

a. as temperature is constant

we need to apply Byle's Law

P1V1=P2V2

where P1 = initial pressure= 1.01 x 105 Pa

V1 = initial volume = 800 m3

P2 = final pressure= 2.01 x 105 Pa ( as gauge pressure is 1)**

V2 = final volume = ?

substituting the values, V2 = 400 m3

Volume of each cylinder = 1.95 m3

so no of cylinders required = 400/1.95= 205.12 or 206

** the question looks incomplete. I feel this is the procedure to solve it.

b. mass of displaced air = density x volume

= 1.23 x 800 = 984 kg

Now, we need to find the mass of hydrogen

For 1 mole of hydrogen at 15 C (i.e. 273 +15=288 K), applying ideal gas equation

PV=nRT

so, 1.01 x 105 x M/d = 1 x 8.3145 x 288 ( V=volume= mass/density=M/d)

so, 1.01 x 105 x 2.02 x 10-3 /d = 1 x 8.3145 x 288

so, d= density of hydrogen at 15 c = 0.08435 kg/m3

so mass of Hydrogen = Volume x density = 800 x 0.08435 = 67.48 kg

applying laws of buoyancy

the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air = 984 - 67.48 = 916.52 kg

c. mass of displaced air = density x volume

= 1.23 x 800 = 984 kg

Now, we need to find the mass of helium

For 1 mole of helium at 15 C (i.e. 273 +15=288 K), applying ideal gas equation

PV=nRT

so, 1.01 x 105 x M/d = 1 x 8.3145 x 288 ( V=volume= mass/density=M/d)

so, 1.01 x 105 x 4 x 10-3 /d = 1 x 8.3145 x 288 ( assuming molar mass of helium=4g/mol)

so, d= density of hydrogen at 15 c = 0.1687 kg/m3

so mass of Hydrogen = Volume x density = 800 x 0.1687 = 134.96 kg

applying laws of buoyancy

the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air = 984 - 134.96 = 849.04 kg

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