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Course: MCAT > Unit 3

Lesson 1: Foundation 4: Physical and chemical principles

Gases: Charles' law and gas in a piston

Problem

According to the experiments of Jacques Charles, when a gas within a closed system is heated, the volume increases proportionally, so long as the pressure and moles for the system remain constant. Charles’s law is represented by the following equation:
VinitialTinitial = VfinalTfinal
*Volume can be expressed in metric units. Temperature must be expressed in kelvins.
The flight of a hot air balloon is attributable to Charles’s law. As the air in the balloon is heated, the volume of the air is increased. In this instance, a given weight of hot air occupies a greater volume than the same given weight of cold air, or said differently, the hot air inside the balloon becomes less dense than the surrounding cooler air. The hot air of the balloon (now less dense) begins to rise, culminating in flight. As the air in the balloon is allowed to cool, the density increases, and the basket returns to the ground.
This law can be demonstrated experimentally by observing the change in volume within a piston. The piston is exposed to heat through an incorporated heating coil, which relies on voltage adjustment to control temperature. Heat is released as a result of increased voltage and subsequent increased current passing through the coil. The following table provides the collected data from one such experience, in which volume change is measured by the height of the internal chamber of the piston. Isobaric conditions apply.
Table 1. A list of corresponding heights and volumes at each temperature.
TemperatureHeightVolume
40oC6.42 cm160.6π cm3
50oC6.42 cm165.9π cm3
60oC6.42 cm171.2π cm3
70oC6.42 cm176.5π cm3
80oC6.42 cm181.8π cm3
90oC6.42 cm187.1π cm3
100oC6.42 cm192.4π cm3
The experimental data adheres closely to ideal gas behavior, and deviations from ideal behavior are negligible.
If the piston became locked in place such that chamber volume remained constant, as voltage in the coil increases, how would the internal pressure affected?
Choose 1 answer: