You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Boyle's Law

A gas occupie...

Question

A gas occupies a volume of $250 c m^{3}$ at $745 t o r r$ and $25^{\circ} C$ . What additional pressure is required to reduce the gas volume to $200 c m^{3}$ at the same temperature?

Open in App

Solution

Using Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
= $250 \times 745 = 200 \times P_{2}$
= $P_{2} = 931.25 t o r r$
Additional Pressure $= 931.25 - 745 = 186.25 t o r r$

Suggest Corrections

Similar questions

Q. A gas occupies a volume of

250

ml at

745

mm Hg and

25^{\circ}

C. What additional pressure is required to reduce the gas volume to

200

ml at the same time?

Q. A gas occupies a volume of

250 m L

745 m m

H g

and

25^{o} C

. What additional pressure is required to reduce the gas volume to

200 m L

at the same temperature?

Q. A sample of gas at

35^{o} C

& 1

a t m

pressure occupies a volume of 3.75 litres. At what temperature should the gas be kept if it is required to reduce the volume to 3 litres at the same pressure?

(in

K

to the nearest integer)

Q. A sample of gas at

35^{o} C

& 1 atm pressure occupies a volume of 3.75litres. At what temp. should the gas be kept if it is required to reduce the volume to 3 litres at the same pressure:

A gas occupies a volume of 2.5 L at $9 \times 10^{5} N m^{- 2}$ . Calculate the additional pressure required to decrease the volume of the gas to 1.51. Assume temperature is constant.

Join BYJU'S Learning Program

A gas occupies a volume of 250 cm3 at 745 torr and 25∘C. What additional pressure is required to reduce the gas volume to 200 cm3 at the same temperature?

Using Boyle's law,P1V1=P2V2=250×745=200×P2=P2=931.25 torrAdditional Pressure=931.25−745=186.25 torr

A gas occupies a volume of $250 c m^{3}$ at $745 t o r r$ and $25^{\circ} C$ . What additional pressure is required to reduce the gas volume to $200 c m^{3}$ at the same temperature?

Using Boyle's law,
$P_{1} V_{1} = P_{2} V_{2}$
= $250 \times 745 = 200 \times P_{2}$
= $P_{2} = 931.25 t o r r$
Additional Pressure $= 931.25 - 745 = 186.25 t o r r$