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 pressure to remain constant?

A. 140 K

B. 273 K

C. 560 K

D. 5600 K

To solve this we assume that the gas inside is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

T2 = T1 x V2 / V1

T2 = 280 x 20.0 / 10

T2 = 560 K