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A certain gas has a volume of 500.0 ml at 77.0c and 600.0 torr. calculate the temperature if the volume decreased to 400.0 ml and the pressure is increased to 760. torr and assume the number of moles does not change.

Answer :

NUHULAWAL20GO

As the volume is decreased and the pressure is increased to the given value, the temperature of the gas increases to 81.67°C.

Given the data in the question;

  • Initial volume of gas; [tex]V_1 = 500.0ml = 0.5L[/tex]
  • Initial pressure; [tex]P_1 = 600.0torr = 0.789474atm[/tex]
  • Initial temperature; [tex]T_1 = 77.0^oC = 350.15K[/tex]
  • Final Volume; [tex]V_2 = 400.0ml = 0.4L[/tex]
  • Final pressure; [tex]P_2 = 760torr = 1.0atm[/tex]
  • Final temperature; [tex]T_2 =\ ?[/tex]

Combined gas law

Combined gas law brings together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

We substitute our given values into the expression above to determine the new temperature.

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\P_1V_1T_2 = P_2V_2T_1\\\\T_2 = \frac{P_2V_2T_1}{P_1V_1}\\T_2 = \frac{1.0atm\ *\ 0.4L\ *\ 350.15K}{0.789474atm\ *\ 0.5L} \\\\T_2 = \frac{140.06LatmK}{0.394737Latm}\\ \\T_2 = 354.82K\\\\T_2 = 81.67^oC[/tex]

Therefore, as the volume is decreased and the pressure is increased to the given value, the temperature of the gas increases to 81.67°C.

Learn more about the combined gas law here: brainly.com/question/25944795

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