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how many moles of gas does it take to occupy 120 liters at a pressure of 2.3 atmospheres and a temperature of 340 K?

Verified Answer :

dexteright02

Data:

P (pressure) = 2.3 atm
V (volume) = 120 L
T (temperature) = 340 K
n (number of mols) = ?
R (Gas constant) = 0.082 (atmL/molK)

Formula: 

\(p*V = n*R*T\)

Solving:
\(p*V = n*R*T\)
\(2.3*120 = n*0.082*340\)
\(276 = 27.88n\)
\(27.88n = 276\)
\(n = \frac{276}{27.88}\)
\(n = 9.89956...\to \boxed{\boxed{n\approx 9.90\:mols}}\end{array}}\qquad\quad\checkmark\)


nuhulawal20

The amount of gas that will occupy the volume at the given pressure and temperature is 9.89moles.

Given the data in the question;

  • Volume of gas; \(V = 120L\)
  • Pressure; \(2.3atm\)
  • Temperature; \(T = 340K\)
  • Amount of gas; \(n = \ ?\)

Ideal Gas Law

The Ideal Gas Law states that "the pressure P multiply by the volume V is equal to moles multiply by temperature and the universal gas constant. It is expressed;

\(PV = nRT\)

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

We substitute our values into the expression above to determine the amount of gas n.

\(PV = nRT\\\\2.3atm * 120L = n * 0.08206Latm/molK * 340K\\\\276Latm = n * 27.9004 Latm/mol\\\ = \frac{276Latm}{27.9004 Latm/mol}\\ \ = 9.89moles\)

Therefore, the amount of gas that will occupy the volume at the given pressure and temperature is 9.89moles.

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

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