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sasho [114]
3 years ago
15

What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperatu

re is held constant?
2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?
Chemistry
1 answer:
tatyana61 [14]3 years ago
3 0

Answer:

1) 6.0 atm.

2) 2.066 atm.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>

  • At constant T and at two different (P, and V):

<em>P₁V₁ = P₂V₂.</em>

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>

<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>

<em></em>

  • Since the container is rigid, so it has constant V.
  • At constant V and at two different (P, and T):

<em>P₁/T₁ = P₂/T₂.</em>

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>

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Cp = 2.12 KJ Kg-1 K-1

Cv = Cp- Ru/M

Cv= 2.12 - 8.314/16.12 = 1.604 KJ Kg-1 K-1

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