You have several identical balloons. You experimentally determine that a balloon will break if its volume...

You have several identical balloons. You experimentally determine that a balloon will break if its volume exceeds 0.820 L. The pressure of the gas inside the balloon equals air pressure (1.00 atm).

(a) If the air inside the balloon is at a constant temperature of 21.8

Expert Solution

(a) PV = nRT...(1)

P = 1 atm, V = 0.820 L, T = 21.8 + 273 = 294.8 K, R = 0.082 L?atm?K^?1?mol^?1,

Molecular mass of air is calculated as follows:

Nitrogen (N2) (78.084%)
Oxygen (O2) (20.946%)
Argon (Ar) (0.9340%)
Carbon dioxide (0.0387%)
Neon (Ne) (0.001818%)
Helium (He) (0.000524%)
Methane (CH4) (0.000179%)
Krypton (Kr) (0.000114%)
& there is more

then add up those %'s of their molar masses:

Nitrogen (N2) (78.084% of 28.0134 g/mol = 21.8740

Oxygen (O2) (20.946%of 31.9989 g/mol = 6.7025

Argon (Ar) (0.9340% of 39.9481 g/mol= 0.3731

Carbon dioxide (0.0387% of 44.0096 g/mol = 0.0170

& so far you get 28.967 g/mol

Substitute all values in equation 1

1 * 0.820 = (w / 28.967) * 0.082 * 294.8

0.820 * 28.967 = w * 241.73

23.75 / 241.73 = w

w = 0.09 g

(b) If helium is there then Molecular mass = 4.0 g

1 * 0.820 = (w / 4.0 ) * 0.082 * 294.8

0.820 * 4.0 = w * 241.73

3.28 / 241.73 = w

w = 0.01 g

queen_honey_blossom answered 2 months ago

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