## 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