A container with a volume of #8 L# contains a gas with a temperature of #370^o C#. If the temperature of the gas changes to #510 ^o K# without any change in pressure, what must the container's new volume be?

Apply Charles' Law

The final volume is

Using the ideal gas law, ( PV = nRT ), where ( P ) is the pressure, ( V ) is the volume, ( n ) is the number of moles of gas, ( R ) is the ideal gas constant, and ( T ) is the temperature in Kelvin, we can rearrange the equation to solve for the new volume:

[ V_2 = \frac{nRT_2}{P} ]

Since the pressure is constant, it cancels out from both sides of the equation.

Given:

• Initial volume ( V_1 = 8 ) L
• Initial temperature ( T_1 = 370 ) °C
• Final temperature ( T_2 = 510 ) K

To solve for the new volume (( V_2 )), we need to convert the initial temperature to Kelvin:

[ T_1 = 370 + 273 = 643 , \text{K} ]

Now, we can plug in the values:

[ V_2 = \frac{nRT_2}{P} = \frac{nR \times T_2}{P} ]

Since the number of moles of gas (n) and the ideal gas constant (R) are constant, they can be combined into a single constant, ( k ):

[ V_2 = k \times T_2 ]

Now, calculate ( k ):

[ k = \frac{nR}{P} ]

Substitute the known values:

[ k = \frac{n \times R}{P} = \frac{8.314 , \text{J/mol K} \times 643 , \text{K}}{101.3 , \text{kPa}} ]

[ k ≈ 52.63 , \text{L kPa/K} ]

Finally, find ( V_2 ):

[ V_2 = k \times T_2 = 52.63 , \text{L kPa/K} \times 510 , \text{K} ]

[ V_2 ≈ 26853 , \text{L kPa} ]

Therefore, the container's new volume must be approximately 26,853 L.