# A bicycle pump contains 50 cm^3 of air at a pressure of 1x10^5 Pa. What would be the volume of the air if the pressure was increased to 2.1x10^5 Pa.. at constant temperature? This problem deals with Boyle's Law which shows the relationship between the pressure of a gas and the corresponding volume. The relationship between pressure and volume is inverse, meaning that when the pressure increases the volume decreases while if the pressure decreases the volume increases. If you remember this relationship,...

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This problem deals with Boyle's Law which shows the relationship between the pressure of a gas and the corresponding volume. The relationship between pressure and volume is inverse, meaning that when the pressure increases the volume decreases while if the pressure decreases the volume increases. If you remember this relationship, you can look at any problem and predict whether the missing variable will be higher or lower as a way checking your answser.

The pertinent formula is:  P1V1 = P2V2

Where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure and V2 is the final volume.

V1 = 50 cc, P1 = 100 kPa

V2 = unknown, and P2 = 210 kPa.

I have reexpressed the pressure in units of kilopascals since that is the standard unit used.

50 x 100 = 210 V2

V2 = 23.8 cc

Approved by eNotes Editorial Team The bicycle pump contains 50 cm^3 of air at a pressure of 1x10^5 Pa. We need to find the volume if the pressure is increased to 2.1x10^5 Pa.

We use the ideal gas law here. PV = nRT, where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature.

Now in both the cases, the temperature is constant, number of moles is constant, and the gas constant is the same.

Therefore PV = nRT => R = PV/nT

50*1x10^5/ nT =  2.1x10^5 * V /nT

=> 50*1x10^5 =  2.1x10^5 * V

=> V = 50*1x10^5 / 2.1x10^5

=> V = 50 * 1 / 2.1 = 23.80 cm^3.

The volume of air with the altered conditions is 23.80 m^3.

Approved by eNotes Editorial Team