Question

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.

Answer 1

Answer:

Heating this gas to 55 °C will raise its volume to 6.87 liters.

Assumption: this gas is ideal.

Explanation:

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

Alternatively, consider the ideal gas law:

$\displaystyle V = \frac{n \cdot R}{P}\cdot T$.

• $n$ is the number of moles of particles in this gas. $n$ should be constant as long as the container does not leak.
• $R$ is the ideal gas constant.
• $P$ 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,

$V\propto T$.

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$.

For the gas in this question:

• Initial volume: $V_1 = \rm 6.24\; L$.

Convert the two temperatures to degrees Kelvins:

• Initial temperature: $T_1 = \rm 25.0\;\textdegree{C} = (25.0 + {\rm 273.15})\; K = 298.15\;K$.
• Final temperature: $T_1 = \rm 55.0\;\textdegree{C} = (55.0 + {\rm 273.15})\; K = 328.15\;K$.

Apply Charles's Law:

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