Presentation on theme: "Heat Flow. Constant Volume Fixing the piston keeps the volume constant. If heat flows in then temperature remains the same. heat flows at base to."— Presentation transcript:
Constant Volume Fixing the piston keeps the volume constant. If heat flows in then temperature remains the same. heat flows at base to change temperature hold fixed
Isochoric Process For a constant volume process there is no work. V = 0 V = 0 W = 0 W = 0 The internal energy change is only due to heat. U = Q – W = Q U = Q – W = Q P V isotherms
Specific Heat at Constant Volume At constant volume the heat equals the change in internal energy. A molar specific heat at constant volume relates to the change in temperature. C V can be defined from the internal energy.
Constant Pressure Allow the piston to move to keep the pressure constant. Same on both sides Heat flows in and the piston can do work. heat flows at base to change temperature move to maintain pressure
Isobaric Process For a constant pressure process the work is a simple product. W = P(V 2 – V 1 ) W = P(V 2 – V 1 ) The heat can be related to the internal energy change and volume change. Q = U + WQ = U + W Q = U + P VQ = U + P V P V isotherms
Specific Heat at Constant Pressure The internal energy was related to the temperature change. A molar specific heat at constant pressure relates to the change in temperature. For and ideal gas, C P can be defined from C V.
Specific Heat for Gases The ideal gas law predicts a simple relationship between the two forms of specific heat. C P = C V + RC P = C V + R R = 1.99 cal/mol-KR = 1.99 cal/mol-K The table shows how close real gases are to ideal gas behavior. Gas C V C P (cal/mol-K) He2.984.97 Ne2.984.97 N 2 4.966.95 O 2 5.037.03 CO 2 6.808.83 H 2 O6.208.20 C 2 H 6 10.3012.35
No Heat Completely insulate the system. Allow the piston to move. Heat can’t flow, but work can be done. Equivalent process occurs when change is quick so little heat flows. completely insulate to block heat flow
Adiabatic Process For an adiabatic process there is no heat flow. U = Q – W = – W U = Q – W = – W There a relationship that can be derived for ideal gases. P V isotherms next