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ohaa [14]
3 years ago
10

The volume of a Goodyear Blimp is 2.5x10^7 L, which is occupied by 1.7x10^6 mol of helium. What is the internal pressure of the

blimp at 22C
Chemistry
1 answer:
attashe74 [19]3 years ago
4 0

Answer:

1.7 bar

Explanation:

We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.

pV = nRT     Divide both sides by V

 p = (nRT)/V

Data:   n = 1.7 × 10⁶ mol

R = 0.083 14 bar·L·K⁻¹mol⁻¹

T = 22 °C

V = 2.5 × 10⁷ L

Calculations:

(a) <em>Change the temperature to kelvins </em>

T = (22 + 273.15) K

  = 295.15 K

(b) Calculate the pressure

p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)

  = 1.7 bar

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K = °C + 273 A 4.1 L sample of gas is held at 25 °C. If the gas expands to 6.8 L, what is the final temperature?
gregori [183]

Answer:

221 °C

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 4.1 L

Initial temperature (T₁) = 25 °C

= 25 °C + 273

= 298 K

Final volume (V₂) = 6.8 L

Final temperature (T₂) =?

The final temperature of the gas can be obtained as follow:

V₁ / T₁ = V₂ / T₂

4.1 / 298 = 6.8 / T₂

Cross multiply

4.1 × T₂ = 298 × 6.8

4.1 × T₂ = 2026.4

Divide both side by 4.1

T₂ = 2026.4 / 4.1

T₂ ≈ 494 K

Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:

°C = K – 273

K = 494

°C = 494 – 273

°C = 221 °C

Thus the final temperature of the gas is 221 °C

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The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What
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Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

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on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

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The heat of vaporization at 75° C = 900 J/mol

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