New Volume
Consider 5 liters of an ideal gas is at a temperature of 400 K. If the temperature decreases to 200 K, what is the gas’s new volume?
Expand Hint
For an ideal gas:
$$$\frac{P_1v_1}{T_1}=\frac{P_2v_2}{T_2}$$$
where
$$P$$
is the pressure,
$$v$$
is the specific volume, and
$$T$$
is the temperature.
Hint 2
If a variable is not mentioned in the problem statement, assume it is constant for an ideal gas equation.
For an ideal gas:
$$$\frac{P_1v_1}{T_1}=\frac{P_2v_2}{T_2}$$$
where
$$P$$
is the pressure,
$$v$$
is the specific volume, and
$$T$$
is the temperature. If a variable is not mentioned in the problem statement, assume it is constant for an ideal gas equation.
$$$\frac{v_1}{T_1}=\frac{v_2}{T_2}$$$
Therefore,
$$$\frac{5L}{400K}=\frac{v_2}{200K}$$$
$$$v_2=\frac{5L(200K)}{400K}=2.5\:L$$$
2.5 L
Time Analysis
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Similar Problems from FE Section: PVT Behavior
179. PVT Behavior of an Ideal Gas
183. Specific Weight
253. Ideal Gas
257. Nozzle Count
513. Nozzles
555. Ideal Volume