Question

A piston-cylinder device contains 0.5 m³ air at 100 kPa and 1000 K. Because the outside is cooler (at 300 K), the hot air starts losing heat to the surroundings and the volume of air starts decreasing. Over a certain period of time the volume is observed to shrink by 50%.

The final temperature is lower or higher?

Determine the boundary work transfer (absolute value). in KJ

Mark the correct statement that takes into account the sign of heat and work transfer.
W pos Q neg only W neg only Q pos W neg Q pos one negative both negative only W pos both positive

Answer 1

According to the thermodynamics conditions the hot air contained in the volume of cylendre under some pressure will have the heat energy stored in the gas (air). The relation of this heat energy is given by W=PV. Where W is the heat energy, P is the pressure and V is the volume of gas enclosed. Assuming that the pressure remains constant during contraction.

According to the given data,

Initial volume = 0.5 m³, Final volume= 50% of Initial volume = 0.5/2 = 0.25 m³

Initial pressure = 100 kilo pascal, Final pressure = 100 kilo pascal

Initial temperature = 1000K And Final temperature = 300K

Using the bove relation of heat energy, W = - PV The negative sign indicates that the heat is transfered to the surrounding.

W = (10⁵)(0.5-0.5/2)

solving for W, we get, W= -2.5 *10⁴ joules

This is the amount of heat energy which is lost (released) to the surrounding when the hot gas cools down. Since the heat energy is lost to the surrounding, this is considered to be negative. The final temperature is also reduced.

Also note that both the Heat energy Q and the work done W are negative in this case because both leave the system.