Boyle's Law Calculator

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**Step 1 :**

**Step 2 :**

The state of a gas is a function of its pressure, volume and temperature. Boyle's law is one of the gas laws that shows the variation of pressure with volume, keeping temperature constant.

It states that:At a given temperature, the pressure of a given mass of a gas is inversely proportional to its volume.

If P is the pressure of a certain mass of a gas and V is its volume at a temperature T Kelvin then

P $\alpha$ $\frac{1}{V}$ ( T constant )

P = $\frac{k}{V}$

PV = k .

Where k is a constant for a given mass of a gas and at a given temperature. k changes with the mass of the gas and its temperature.

Suppose the gas has Initial pressure as P_{i} and Volume V_{i}. If pressure is increased or decreased the Pressure and Volume changes to P_{f} and V_{f}. Then Boyles law can also be given as

P_{i}V_{i} = P_{f}V_{f} = Constant

Where P_{i} = Initial Pressure

P_{f} = Final Pressure

V_{i }= Initial Volume

V_{f} = Final Volume.

P

V

V

Boyle's law Calculator helps to determine the Unknown quantity among the two i.e Pressure or Volume When there is change in either pressure or Volume from its initial state.

Analyze what are the parameters given in the problem and what is the Unknown quantity.

Using the formula

P_{i} V_{i} = P_{f} V_{f}

Where P_{i} = Initial Pressure

P_{f} = Final Pressure

V_{i} = Initial Volume

V_{f} = Final Volume.

Substituting the known any of the three quantities in the above formula, we get the Unknown quantity.

If a gas occupies 76ml at 760 mm pressure.Using boyle's Law, determine the volume of the gas at the same temperature but at 76 mm pressure?

**Step 1 :**Given: P

_{i}= 760 mm,

V

_{i}= 76 ml,

P

_{f}= 76 mm

V

_{f}= ?**Step 2 :**P

_{i}V_{i}= P_{f}V_{f}

V

_{f}= $\frac{P_{i} V_{i}}{P_{f}}$

= $\frac{760 mm \times 76 ml}{ 76 mm}$

**Answer :**0.760 litres.

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L. If the balloon is filled with 2.0 L of helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only 500 mm Hg, will the balloon burst? (Assume temperature to be constant).

**Step 1 :**At sea level,

P1 = 1 atm = 760 mm of Hg P2 = 500 mm of Hg

V1 = 2 L V2 =?**Step 2 :**Using formula, P

_{1}V_{1}= P_{2}V_{2}

V

_{2}= $\frac{P_{1} \times V_{1}}{P_{2}}$

= $\frac{760 \times 2}{500}$**Answer :**3 Litres