If the temperature is kept constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one-third of its original value?

Using the ideal gas equation pV = nRT, where R is the ideal gas constant, we can show that p=nRT/V. Since n, R and T are all constants, p2/p1 = V1/V2 where p1 and p2 are the start and final pressures respectively and V1 and V2 are the start and final volumes respectively. For if p1 = 3*p2, the pressure would have fallen to one third of its original value, and it follows that V2 = 3*V1. Therefore, for the pressure to fall to a third of its original value, the volume must increase by a factor of 3.