Question

At a temperature of 320K, the gas in a cylinder has a volume of 40.0 liters. If the volume of the gas is decreased to 20.0 liters, what must the temperature be for the gas pressure to remain constant?

A. 160 K
B. 273 K
C. 560 K
D. 140 K

Answer 1

The correct answer is option A.

Step-by-step explanation:

Initial volume of the gas
`V_1= 40.0L`

Initial temperature of gas
`T_1= 320 K`

Final volume of the gas
`V_2= 20.0L`

Final temperature of the gas =
`T_2`

Applying Charles' Law:

`(V_1)/(T_1)=(V_2)/(T_2)`

`T_2=(V_2* T_1)/(V_1)=(20.0L* 320 K)/(40.0L)=160K`

The temperature of the gas when volume of the gas is 20.0 L is 160 K.Hence, the correct answer is option A.

Answer 2

Assuming the gas behaves ideally,
PV/T = constant. P will also be constant in this giving us:
V₁/T₁ = V₂/T₂
40/320 = 20/T₂
T₂ = 160 K
The answer is A.