At a temperature of 280 K, the gas in a cylinder has a volume of 20.0 liters. If the volume of the gas is decreased to 10.0 liters, what must the temperature be for the gas to remain at a constant pressure?

PV=nRT

P is Pressure (#Pa# or Pascals)
V is Volume (#m^3# or metres cubed)
n is Number of moles of gas (#mol# or moles)
R is the Gas constant (#8.31 JK^-1mol^-1# or Joules per Kelvin per mole)
T is Temperature (#K# or Kelvin)

In this problem, you are multiplying V by #10.0/20.0# or #1/2#. However, you are keeping all the other variables the same except T. Therefore, you need to multiply T by 2, which gives you a temperature of 560K.

PV=nRT

P is Pressure (Pa or Pascals)
V is Volume (m3 or metres cubed)
n is Number of moles of gas (mol or moles)
R is the Gas constant (8.31JK−1mol−1 or Joules per Kelvin per mole)
T is Temperature (K or Kelvin)

Since we are only looking at 2 variables (temperature and volume) you can remove all other parts of the equation. That leaves you with

V=T

Therefore is volume drops in half then temperature drops in half to keep the pressure the same.
This makes sense if you think about a balloon. If you squish it down to half it's size the pressure will go up. The only way to reduce the pressure would be to drop the temperature.