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#### Gas Laws:

- Mathematical relationships between volume, pressure, and temperature of a given mass of a gas are referred as Gas laws.

#### Boyle’s Law:

- In 1662, Robert Boyle, on the basis of his experiments put forward the law. It is the relation between the volume and the pressure of enclosed gas at constant temperature.
**Statement:**At constant temperature, the volume of a certain mass of enclosed gas varies directly with its pressure.**Explanation**: Let P be the pressure and V be the volume of a certain mass of enclosed gas, then at the constant temperature

P ∝ 1/V

Thus PV = Constant.

- Thus, for a given amount of the gas, the product of pressure and volume is constant at constant temperature.
- If V
_{1 }and V_{2}are the volumes of a gas at pressures P_{1}and P_{2}respectively at a constant temperature.

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

This relation is called the mathematical statement of Boyle’s Law.

#### Graphical representation :

#### Graph of Pressure (P) Versus Volume (V):

- A graph is drawn by taking volume on the x-axis and pressure on the y-axis. The graph is as follows. This graph is also known as PV diagram.

- Each curve is rectangular hyperbola and corresponds to a different constant temperature and is known as an isotherm (constant temperature plot). Higher curves correspond to higher temperature.
- It should be noted that volume of the gas doubles if pressure is halved.

#### Graph of Pressure (P) Versus Reciprocal of Volume (1 / V):

- The straight lines obtained in the graph confirms that P ∝ 1/V.

#### Graph of Product of Pressure and Volume (PV) Versus Pressure (P):

- The graph parallel to x-axis confirms that at a particular temperature of the gas, the product of its volume and corresponding pressure is always constant.

#### Graphs in terms of Logarithmic Variations:

By Boyle’s law, we have PV = k = constant

Taking log of both sides

Log P + Log V = log K

∴ LogP = – LogV + log K

∴ LogP = log (1/V) + log k

#### Relation Between Density and Pressure:

Thus the density of the certain mass of an enclosed gas is directly proportional to its pressure.

### Problems on Boyle’s Law:

#### Example – 01:

- 5 dm
^{3}volume of a gas exerts a pressure of 2.02 × 10^{5}kPa. This gas is completely pumped into another tank where it exerts a pressure of 1.01 × 10^{5}kPa at the same temperature, calculate the volume of the tank **Solution:****Given:**Initial volume = V_{1}= 5 dm^{3}, Initial pressure = P_{1}= 2.02 × 10^{5}kPa, Final pressure P_{2}= 1.01_{1}kPa, Temperature constant**To Find:**Final volume = V_{2}= ?

At constant temperature by Boyle’s law

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

∴ V_{2} = P_{1}V_{1} /P_{2}

∴ V_{2} = 2.02 × 10^{5} kPa × 5 dm^{3} / 1.01 × 10^{5} kPa

∴ V_{2} = 10 dm^{3}

Hence the volume of the tank is 10 dm^{3}.

#### Example – 02:

- Given the mass of a gas occupies a volume of 2.5 dm
^{3}at NTP. calculate the change in volume of gas at same temperature if the pressure of the gas is changed to 1.04 × 10^{5}Nm^{-2}.

**Solution:****Given:**Initial volume = V_{1}= 2.5 dm3, Initial pressure = P_{1}= 1.013 × 10^{5}Nm^{-2}(Normal pressure), Final pressure P_{2}= 1.04 × 10^{5}Nm^{-2}, Temperature constant**To Find:**Final volume = V_{2}= ?

At constant temperature by Boyle’s law

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

∴ V_{2} = P_{1}V_{1} /P_{2}

∴ V_{2} = 1.013 × 10^{5} Nm^{-2} × 2.5 dm^{3} / 1.04 × 10^{5} Nm^{-2}

∴ V_{2} = 2.435 dm^{3}

∴ Change in volume = V_{1} – V_{2} = 2.5 dm^{3} – 2.435 dm^{3} = 0.065 dm^{3}

Hence the change in volume of the gas is 0.065 dm^{3}.

#### Example – 03:

- A balloon is inflated with helium gas at room temperature of 25 C and at 1 bar pressure when its initial volume is 2.27 L and allowed to rise in the air. As it rises the external pressure decreases and volume of the gas increases till finally, it bursts when external pressure is 0.3 bar.What is the limit to which volume of the balloon can be inflated?
**Solution:****Given:**Initial volume = V_{1}= 2.27 L, Initial pressure = P_{1}= 1 bar, Final pressure P_{2}= 0.3 bar, Temperature constant**To Find:**Final volume = V_{2}= ?

At constant temperature by Boyle’s law

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

∴ V_{2} = P_{1}V_{1} /P_{2}

∴ V_{2} = 1 bar × 2.27 L/ 0.3 bar

∴ V_{2} = 7.567 L

Thus balloon can be inflated to the maximum volume of 7.567 L.

#### Example – 04:

- The volume of given mass of a gas is 0.6 dm
^{3 }at a pressure of 101.325 kPa. Calculate the volume of the gas if its pressure is ballooned to 142.860 kPa at the same temperature.

**Solution:**

**Given:**Initial volume = V_{1}= 0.6 dm^{3}, Initial pressure = P_{1}= 101.325 kPa, Final pressure P_{2}= 142.860 kPa, Temperature constant

**To Find:**Final volume = V_{2}= ?

At constant temperature by Boyle’s law

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

∴ V_{2} = P_{1}V_{1} /P_{2}

∴ V_{2} = 101.325 kPa × 0.6 dm^{3} / 142.860 kPa

∴ V_{2} = 0.426 dm^{3}

Thus final volume of the gas is 0.426 dm^{3}.

#### Example – 05:

- In a J tube partially filled with mercury, the volume of the air column is 4.2 mL and the mercury level in the two limbs is the same. Some mercury is now added to the tube so that the volume of air enclosed in the shorter limb is now 2.8 mL. What is the difference in the level of mercury in this case? Atmospheric pressure is 1 bar

**Solution:**

**Given:**Initial volume = V_{1}= 4.2 mL, Initial pressure = P_{1}= 1 bar, Final volume V_{2}= 2.8 mL Temperature constant

**To Find:**Difference in mercury levels =?

At constant temperature by Boyle’s law

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

∴ P_{2} = P_{1}V_{1} /V_{2}

∴ P_{2} = 1 bar × 4.2 mL / 2.8 mL

∴ P_{2} = 1.5 bar

Difference in pressure = P2 – P1 = 1.5 bar – 1 bar = 0.5 bar

Now 1 bar corresponds to 750.12 mm of mercury

Hence 0.5 bar corresponds to 750.12 x 0.5 =375.06 mm of mercury

Hence the difference in mercury levels is 375.06 mm or 37.51 cm

#### Example – 06:

- A thin glass bulb of 100 mL capacity is evacuated and kept in 2.0 L container at 27 °C and 800 mm pressure. If the bulb implodes isothermally, calculate the new pressure in the container in kPa.
**Solution:****Given:**Initial Volume = 2000 mL – 100 mL = 1900 mL**To Find:**Final pressure = ?

At constant temperature by Boyle’s law

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

∴ P_{2} = P_{1}V_{1} /V_{2}

∴ P_{2} = 800 mm × 1900 L / 2000 mL

∴ P_{2} = 76o mm = 760 mm/760 mm = 1 atm = 101.325 kPa

Hence the new pressure is 101.325 kPa

#### Example – 07:

- Certain bulb A containing gas at 1.5 bar pressure was put into communication with an evacuated vessel of 1.0 dm
^{3}capacity through the stop cock. The final pressure of the system dropped to 920 mbar, at the same temperature. What is the volume of container A. **Solution:****Given:**Initial pressure P1 = 1.5 bar, Final Pressure = 920 mbar = 0.920 bar. Let V dm^{3 }be the volume of the container A. Initial Volume = V1 = V dm^{3}, Final Volume = (1.0 + V) dm^{3}**To Find:**Volume of container A = V = ?

At constant temperature by Boyle’s law

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

∴ 1.5 bar × V dm^{3} = 0.920 bar × (100 – V) dm^{3}

∴ 1.5 V = 92 – 0.920V

∴ 1.5 V + 0.920V = 92

∴ 2.42 V = 92

∴ V = 92 / 2.42 = 39.02

Hence the volume of container A is 38.02 dm³.

#### Example – 08:

- What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30 °C
**Given:**Initial volume = V_{1}= 500 dm³, Initial pressure = P_{1}= 1 bar, Final volume V_{2}= 200 dm³ Temperature constant**To Find:**Final Pressure = P_{2}= ?

At constant temperature by Boyle’s law

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

∴ P_{2} = P_{1}V_{1} /V_{2}

∴ P_{2} = 1 bar × 500 dm³ / 200 dm³

∴ P_{2} = 2.5 bar

Hence minimum pressure required = 2.5 bar

#### Example – 09:

- A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be the pressure?

Ans: 0.8 bar

**Given:**Initial volume = V_{1}= 120 mL, Initial pressure = P_{1}= 1.2 bar, Final volume V_{2}= 180 mL. Temperature constant**To Find:**Final Pressure = P_{2}= ?

At constant temperature by Boyle’s law

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

∴ P_{2} = P_{1}V_{1} /V_{2}

∴ P_{2} = 1.2 bar × 120 dm³ / 180 dm³

∴ P_{2} = 0.8 bar

Hence minimum pressure required = 0.8 bar

Boyle’s law Boyle’s law Boyle’s law Boyle’s law Boyle’s law Boyle’s law Boyle’s law Boyle’s law Boyle’s law Boyle’s law

Science > Chemistry > States of Matter > You are Here |

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I accept P inversely proportion to V . However if we check Volume at different location having different temperature i.e. T1 , T2 & T3 , Is Volume w.r.t. Pressure will give same result ?

EXP. If compressor gives flow 100 Lit / Sec at 7 bar pressure in Europe where we considered Ambient temp is 7deg cel. If same compressor run in India having Amb temp is 40 Deg Cel. In such condition how much flow it will delivered w.r.t. same pressure i.e 7 Bar. Is Flow will vary as Amb temperature has changed. If it would vary. what would be flow in Lit / Sec

Pl explain the Limitation of Bayle Law with examples.