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?

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