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calculate the pressure in atmospheres of 1.92 moles of helium gas in a 10.0L container at 25C

Sagot :

The pressure in atmospheres of the given mole and volume of helium gas is 4.7atm.

Given the data in the question;

  • Amount of helium; [tex]n = 1.92mol[/tex]
  • Volume of the gas; [tex]V = 10.0L[/tex]
  • Temperature; [tex]T = 25^0C = 298.15K[/tex]

Pressure in atmosphere; [tex]P = \ ?[/tex]

Ideal Gas Law

The Ideal gas law emphasizes on the behavior of a hypothetical ideal gas. It states that pressure P multiply by volume V is equal to moles times temperature and the universal gas constant.

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]R = 0.08206Latm/molK[/tex] ).

To determine the pressure in atmosphere, we substitute our values into the expression above.

[tex]PV = nRT\\\\P = \frac{nRT}{V} \\\\P = \frac{1.92mol\ *\ 0.08206Latm/molK\ *\ 298.15K}{10.0L} \\\\P = \frac{46.975Latm}{10.0L}\\\\P = 4.7atm[/tex]

Therefore, the pressure in atmospheres of the given mole and volume of helium gas is 4.7atm.

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