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laiz [17]
3 years ago
8

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 Questio

n 18 options: 2.84 L. 6.87 L. 13.7 L. 5.67 L.
Chemistry
1 answer:
Sphinxa [80]3 years ago
7 0

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.

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