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docker41 [41]
3 years ago
15

At constant temperature, a sample of helium at 760. torr in a closed container was compressed 5) from 5.00 L to 3.00 L, with no

change in amount of gas or temperature. What was the new pressure exerted by the helium on its container?
A) 3820 torr
B) 1270 torr
C) 800.torr
D)15.0 torr
E) 2280 torr
Chemistry
1 answer:
jek_recluse [69]3 years ago
5 0

Answer:

B) 1270 torr

Explanation:

Given data

  • Initial volume (V₁): 5.00 L
  • Initial pressure (P₁): 760 torr
  • Final volume (V₂): 3.00 L
  • Final pressure (P₂): ?

We can find the final pressure using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁/V₂

P₂ = 760 torr × 5.00 L/3.00 L

P₂ = 1.27 × 10³ torr = 1270 torr

The final pressure is 1270 torr.

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You have 4 moles of a gas in a 50 L container held at 2 atm pressure. Currently the temperature is 27 ºC. R = 0.0821 L*atm/(mol*
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<span>• What is the temperature in Kelvins?
</span>You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15= 
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It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.
</span>
<span>• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
</span>In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
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<span> If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
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This problem is to use the Claussius-Clapeyron Equation, which is:

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ΔH is the enthalpy of vaporization

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T2 and T1 are the temperatures at the estates 2 and 1.

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Then p2 = 101.325 kPa
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