## 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 gasâ€™s new volume?

##
__
__**Hint**

**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**

**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