Pressure - Volume Graph

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Presentation transcript:

Pressure - Volume Graph T4 T3 Isotherms (lines of constant temperature) T2 P T1 Area under curve represents work Pressure Internal energy is proportional to temperature Volume V

Thermodynamic Processes A process which involves heat transfer and/or temperature change and/or work… A. Isobaric B. Isochoric / Isovolumetric C. Isothermal D. Adiabatic

Isobaric (Constant Pressure) Expansion W = -PDV Po W = -ve DU increases T4 T3 T2 1st Law: DU = Q + W T1 DV V ΔV = +ve

Isobaric (Constant Pressure) Compression W = -PDV P Po W = +ve T4 DU decreases T3 T2 1st Law: DU = Q + W T1 DV V ΔV = -ve

Isovolumetric or Isochoric (Constant Volume) P W = 0 Decrease in Pressure Po Pf DU decreases 1st Law: Q = DU T3 T2 T1 DU = Q + W V

Isovolumetric or Isochoric (Constant Volume) P Increase in Pressure W = 0 Po Pf DU increases 1st Law: Q = DU T3 T2 T1 DU = Q + W V

Isothermal (Constant Temperature)

Isothermal (Constant Temperature) The work W done by the gas is not given by W=PΔV =P(Vf -Vi ) because the pressure is not constant. Nevertheless, the work is equal to the area under the graph. The techniques of integral calculus lead to the following result for W: You DO NOT need to know this for AP Physics 2… P Calculus!

Isothermal (Constant Temperature) Expansion P Po so W is -ve (ie gas is expanding) T3 Pf DU = 0 T2 1st Law: DU = Q+W 0 = Q+W Q = -W = (-)(-) = +ve! i.e. the work done is provided by Q added T1 Vo Vf V

Isothermal (Constant Temperature) Compression Po Pf Vo Vf P so W is +ve (ie gas is compressed) T3 DU = 0 T2 1st Law: DU = Q+W 0 = Q+W Q = -W = (-)(+) = ve! T1 V

Isothermal Example How much work does it take to compress 2.5 mol of an ideal gas to half its original volume while maintaining a constant 300 K temperature? P ln(0.5) ~ (-ve) W is +ve as gas is compressed…

The state of an ideal gas was changed three times at three different temperatures. The diagram represents three different isothermal curves. Which of the following is true about the temperature of the gas? T1 < T2 < T3 T1 > T2 > T3 T1 > T2 < T3 T1 > T2 = T3

Adiabatic (No Heat Exchange)

Adiabatic (No Heat Exchange) P Po Pf Vo Vf W = DU Expansion W = -ve DU = decreases T2 1st Law: Q = 0 DU = W T1 V

Adiabatic (No Heat Exchange) P Po Pf Vo Vf W = DU Compression W = +ve DU = increases T2 1st Law: Q = 0 DU = W T1 V

A gas trapped in a container with a movable piston undergoes an adiabatic process. During this process, the gas does about 1025 joules of work. Which of the following statements are true? (select 2 answers) The internal energy of the system increases by 1025 J The temperature of the gas increases The temperature of the gas decreases The heat transferred in or out of the system is zero

MC Questions that follow will test your understanding of all 4 thermal processes…