Ideal Volume
Consider 10 liters of an ideal gas is at a temperature of 300 K. If the temperature decreases to 250 K, what is the gas’s new volume (L)?
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,
$$$\frac{10L}{300K}=\frac{v_2}{250K}$$$
$$$v_2=\frac{10L(250K)}{300K}=8.3\:L$$$
8.3 L
Time Analysis
See how quickly you looked at the hint, solution, and answer. This is important for making sure you will finish the FE Exam in time.- Hint: Not clicked
- Solution: Not clicked
- Answer: Not clicked
Similar Problems from FE Sub Section: Ideal Gas
179. PVT Behavior of an Ideal Gas
183. Specific Weight
253. Ideal Gas
257. Nozzle Count
288. New Volume
513. Nozzles
Similar Problems from FE Section: PVT Behavior
179. PVT Behavior of an Ideal Gas
183. Specific Weight
253. Ideal Gas
257. Nozzle Count
288. New Volume
513. Nozzles