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Kobotan [32]
3 years ago
9

What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?

Chemistry
1 answer:
deff fn [24]3 years ago
3 0

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

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Given that:

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replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

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Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

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Using the relation for the number of moles which is:

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