# Numerical Problem, boyle’s law, For Class 9, ICSE

Hello my students, today we discussing some **numerical application on Gas equation.**

**Prob 1:**

A gas occupies 600 cm^{3 }under a pressure of 700 mm Hg. Find under what pressure the volume of the gas will be reduced by 20 percent of its original volume, The Temperature remaining constant throughout.

Ans:

Given Data: Initial pressure P_{1} = 700 mm Hg, Initial volume V1 = 600 cm^{3}

Final Volume V_{2} = 20 percent of its original volume = ( 600-20 percent of 600) = ( 600 -120) = 480 cm^{3}

P_{2} =?

From Boyels law P_{1} V_{1 }= P_{2} V_{2}

700 X 600 = P_{2} x 480

P_{2} = 875 mm Hg ( Ans)

**Prob 2.**

Two cylinder, both containing carbon di oxide, are connected together by a tube fitted with a tap. The capacity of one cylinder is 4 dm^{3} and that of other 1 dm^{3}, the Pressure in the first cylinder is 560 mm Hg and that in second 1000 mm Hg. What will be the final pressure in both cylinders on opening the Tap, If the temperature remains constant.

Ans: P1 = 560 mm Hg , P_{2} = ?, V_{1} = 4 dm^{3} V_{2} = ( 4+1 ) dm^{3}

Applying boyle’s law P_{1} V_{1 }= P_{2} V_{2}

560 x 4 =P_{2} x 5

P_{2} = 448 mm Hg

2nd case ( for 2nd cylinder )

Applying boyle’s law P_{1} V_{1 }= P_{2} V_{2}

1000 x 1 =P_{2} x 5

P_{2} = 200 mm Hg

Final pressure 448+200 = 648 mm Hg ( Ans )

**Prob 3**

The volume of a given mass of a gas with some pieces of marble in a container at 760 mm of pressure is 100 ml. If the pressure is changed to 1000 mm of Hg, the new volume is 80 ml. Find out the volume occupied by the marble pieces, if the temperature remains constant.

Ans: Let volume occupied by the marble = V ml

at 760 mm Hg ,the volume occupied by gas = ( 100 – V) ml

at 1000 mm Hg ,the volume occupied by gas = ( 80 – V) ml

Applying boyle’s law P_{1} V_{1 }= P_{2} V_{2}

760 x ( 100-V) = 1000 ( 80 – V) solve V

Prob 4

The Pressure of gas A ( P_{A} ) is 3.0 atm when it occupies 5 L of the Volume. Calculate the Final pressure when it is compressed to 3.0L volume at constant Temperature.

Ans: P_{A} = 3.0 atm , Va = 5 ltr , Pb = ? , Vb = 3 ltr

Applying boyle’s law P_{1} V_{1 }= P_{2} V_{2} , Put value and solve Pb

**Writer Identity:** **Sourav Sir** is a well-known Chemistry faculty of Aims ( a well known JEE Main & Advanced and NEET Coaching institute in **Barrackpore**) . He hold **Bachelor of Civil engineering ( BE) & Master of Civil Engineering ( ME) from Jadavpur University, He also did MBA (F) and A Chartered Engineer from Institute of Engineers (I)**. To contact the author please call Aims: +917003557150,+918697234172,+919748932154 Or Email : haldarsourav@gmail.com or log in: http://aimscognitive.com/.