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

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.

Answer 2

Answer:

The correct answer is option A.

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:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$T_2=\frac{V_2\times T_1}{V_1}=\frac{20.0L\times 320 K}{40.0L}=160K$

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