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Stolb23 [73]
3 years ago
9

A sample of ideal gas is in a sealed container. The pressure of the gas is 485 torr , and the temperature is 40 ∘C . If the temp

erature changes to 74 ∘C with no change in volume or amount of gas, what is the new pressure, P2, of the gas inside the container?
Chemistry
1 answer:
Fantom [35]3 years ago
3 0

Answer:

537.68 torr.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and V are constant, and have different values of P and T:

<em>(P₁T₂) = (P₂T₁).</em>

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, ​T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

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<u>Step 1: Define</u>

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<u>Step 2: Solve</u>

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  2. [Pressure] Multiply:                                                                                           8.08 atm · L = P₂(12.0 L)
  3. [Pressure] [Division Property of Equality] Isolate unknown:                          0.673333 atm = P₂
  4. [Pressure] Rewrite:                                                                                           P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

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