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A gas has a volume of 3.00 L and a pressure of 75.4 kPa. When the volume expands to 4.00L and the pressure drops to 72.7 kPa, the gas temperature is 0 o C. What was the initial temperature of the gas?

Respuesta :

Answer:

Initial temperature = 212.35 K or -60.65°C

Explanation:

Boyle - Gay Lussac's Law:

If the amount of particles is constant inside an enclosed container, then the Pressure is inversely proportional to the Volume but direct proportional to the Temperature:

[tex]\boxed{\frac{P_1\cdot V_1}{T_1} =\frac{P_2\cdot V_2}{T_2}}[/tex]

where:

  • P = Pressure
  • V = Volume
  • T = Temperature → (take note that the temperature has to be in Kelvin (K))

Given:

V₁ = 3.00 L

P₁ = 75.4 kPa

V₂ = 4.00 L

P₂ = 72.7 kPa

T₂ = 0°C = 273 K

[tex]\displaystyle\frac{P_1\cdot V_1}{T_1} =\frac{P_2\cdot V_2}{T_2}[/tex]

[tex]\displaystyle\frac{75.4\times10^3(3.00)}{T_1} =\frac{72.7\times10^3(4.00)}{273}[/tex]

[tex]\displaystyle\frac{75.4\times3}{T_1} =\frac{72.7\times4}{273}[/tex]

[tex]\displaystyle T_1 =\frac{75.4(3)(273)}{72.7(4)}[/tex]

[tex]T_1=212.35\ K[/tex]

    [tex]=(212.35-273)^oC[/tex]

    [tex]=-60.65^oC[/tex]

     

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