Question

A balloon contains 7.2 L of He. The pressure is reduced to 2.00 atm and the balloon expands to occupy a volume of 25.1 L. What was the initial pressure exerted on the balloon?

Answer 1

Given that

P1= ?

V1= 7.2 L

P2=2.00atm  

V2= 25.1 L

P1V1= P2V2

P1 = P2V2/V1

= 2.00 atm *25.1 L/ 7.2 L

=6.97 atm

Answer 2

Answer:

Concept:

Boyle's law, also known as Boyle–Mariotte law or Mariotte's law (particularly in France), is an experiment gas law that explains how the pressure of the gas tends to drop as the container volume increases. This is a current version of Boyle's law. If the temperature and amount of gas in a closed system stay constant, the absolute pressure exerted by an unit weight of a thermodynamic equilibrium is inversely proportionalto its volume it occupies.

Given:

A balloon contains 7.2 L of He. The balloons expands to a volume of 25.1 L once the pressure is decreased to 2.00 atm.

Find:

find What was the initial pressure exerted on the balloon?

Answer:

Because pressure and volume have such a direct relationship, if the volume increases, the pressure decreases, and if the pressure increases, the volume decreases (and vice versa) P1V1 = P2V2 is the equation for Boyle's Law. Initial Pressure (P1) (atm or mmHg)

$Initial.volume=V_{1} (L .or. mL)\\Initial.volume.V_{1} =7.2L\\\\Final.pressure.P_{2}=2 atm\\ Final.volume.V_{2} =25.1 L\\Find, Initial .pressure(P_{1} )$

Boyle's Law Equation

$P_{1} V_{1}=P_{2} V_{2}\\P_{1}7.2=2*25.1\\P_{1}7.2=50.2\\P_{1}=\frac{50.2}{7.2} *\frac{10}{10}$

    $=\frac{502}{72} \\\\=\frac{251}{36} \\\\=6.97 atm$

The balloon was first inflated to a initial pressure of $6.97 atm$

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