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navik [9.2K]
3 years ago
11

A same of gas in a rigid container is at 25.0Cand 1.00atm. What is the pressure of the sample when heated to 220.0C

Chemistry
1 answer:
Crazy boy [7]3 years ago
7 0

Answer:

1.654 atm.

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>

<em></em>

  • Knowing that:

P₁ = 1.0 atm, T₁ = 25°C + 273 = 298 K,

P₂ = ??? atm, T₂ = 220°C + 273 = 493 K,

  • Applying in the above equation

<em>(P₁T₂) = (P₂T₁)</em>

<em></em>

<em>∴ P₂ = (P₁T₂)/(T₁) </em>= (1.0 atm)(493 K)/(298 K) = <em>1.654 atm.</em>

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The given equation has been assessed as follows:

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