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What is the temperature of 12.20 mol of gas in a 18.35 l tank at 16.40 atm?

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nuhulawal20

The temperature of the given mole and pressure of gas is 300.6K.

Given the data in the question;

  • Amount of gas; [tex]n = 12.20mol[/tex]
  • Volume of gas; [tex]V = 18.35L[/tex]
  • Pressure; [tex]P = 16.40atm[/tex]

Temperature; [tex]T = \ ?[/tex]

The Ideal Gas Law

The Ideal gas law or general gas equation emphasizes on the state or behavior of a hypothetical ideal gas. It states that  pressure multiplied by volume is equal to moles multiply by the universal gas constant multiply by temperature. It is expressed;

[tex]PV = nRT[/tex]

Where P is pressure, V is volume, n is the amount of substance, T is temperature and R is the ideal gas constant ( [tex]0.08206Latm/molK[/tex] 0

The determine the temperature of the gas, we substitute our given values inro the equation above.

[tex]PV = nRT\\\\T = \frac{PV}{nR}\\ \\T = \frac{16.40atm\ *\ 18.35L }{12.20mol\ * \ 0.08206Latm/molK} \\\\T = \frac{300.94Latm}{1.001132molLatm/molK} \\\\T = 300.6K[/tex]

Therefore, the temperature of the given mole and pressure of gas is 300.6K.

Learn more about Ideal Gas Law: brainly.com/question/4147359

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