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
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