The temperature of 6.24 L of a gas is increased from 25.0°C to 55.0°C at constant pressure. The new volume of the gas is Question 18 options: 2.84 L. 6.87 L. 13.7 L. 5.67 L.

By Charles's Law, under constant pressure the volume [tex]V[/tex] of an ideal gas is proportional to its absolute temperature [tex]T[/tex] (the one in degrees Kelvins.)

Alternatively, consider the ideal gas law:

[tex]\displaystyle V = \frac{n \cdot R}{P}\cdot T[/tex].

[tex]n[/tex] is the number of moles of particles in this gas. [tex]n[/tex] should be constant as long as the container does not leak.
[tex]R[/tex] is the ideal gas constant.
[tex]P[/tex] is the pressure on the gas. The question states that the pressure on this gas is constant.

Therefore the volume of the gas is proportional to its absolute temperature.

Either way,

[tex]V\propto T[/tex].

[tex]\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}[/tex].

For the gas in this question:

Initial volume: [tex]V_1 = \rm 6.24\; L[/tex].

Convert the two temperatures to degrees Kelvins:

Initial temperature: [tex]T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K[/tex].
Final temperature: [tex]T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K[/tex].

Apply Charles's Law:

[tex]\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 6.24\;L \times \frac{328.15\; K}{298.15\;K} = 6.87\;L[/tex].