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...

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.