Question

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

Answer 1

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

Answer 2

Answer is: new temperature is A. 140 K.

V₁(gas) = 20.0 L; initial volume.

T₁(gas) = 280 K.; initial temperature

V₂(gas) = 10.0 L; final volume

T₂(gas) = ?; final temperature.

Charle's Law (the Temperature-Volume Law): the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:

V₁/T₁ = V₂/T₂.

20.0 L/ 280 K = 10.0 L /  T₂.

T₂ = 140 K.

As the volume goes down, the temperature also goes down, and vice-versa.