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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.

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Nuhulawal20idn

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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