Thermodynamics of an Ideal Gas

A State Function Does not depend on how the system arrived at its present state; only on the characteristics of the present state. Volume, Pressure, Temperature, ΔE, ΔH

State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. 2012 Pearson Education, Inc. Thermochemistry

State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. In the system depicted in Figure 5.9, the water could have reached room temperature from either direction. 2012 Pearson Education, Inc. Thermochemistry

State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so, ΔE depends only on E initial and E final. 2012 Pearson Education, Inc. Thermochemistry

State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its ΔE is the same. But q and w are different in the two cases. 2012 Pearson Education, Inc. Thermochemistry

Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). 2012 Pearson Education, Inc. Thermochemistry

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = PΔV 2012 Pearson Education, Inc. Thermochemistry

Ideal Gas The hypothetical condition approached by real gases at high temperatures and low pressures (PV = nrt) The average translational energy for one mole of gas at a given temperature in Kelvins (KE) ave = 3 2 RT

Molar Heat Capacity ( C ) The energy required to raise the temperature of 1mole of a substance by 1K. heat q = n C moles change in temperature

Molar Heat Capacity ( C ) The energy required to raise the temperature of 1mole of a substance by 1K. heat q = n C J/mol K moles change in temperature

Heating an Ideal Gas at constant volume (KE) ave = 3 2 RT 3 (heat energy required) = 2 RT No change in volume - no work done C v = 3 2 R heat energy required to change the temp. of one mole of monoatomic gas 1 K at constant volume

Heating an Ideal Gas at constant pressure the volume increases work is done (heat energy required) = (heat energy required to change the translation energy) + (the energy needed to do the PV work)

Heating an Ideal Gas at constant pressure the volume increases work is done (heat energy required) For a 1 mole 1 K change 3 = R + PΔV 2 PΔV = nr = R C p = 3 2 R + R C p = C v + R heat energy required to change the temp. of one mole of monoatomic gas 1 K at constant pressure

Heating a Polyatomic Gas polyatomic gases absorb energy to excite rotational and vibrational motions in addition to translational motions causing higher C v than (3/2)R C p = C v + R Assuming ideal behavior if C v is known C p can be calculated for any gas

Heating a Gas and Energy ΔE = 3 2 R for n moles ΔE = C v ΔE = nc v At constant pressure work is done (heat energy required) = q p = nc p = n(c v + R) ΔΗ = nc v + nr { ΔE { PΔV

Heating a Gas and Enthalpy Η = E + PV ΔΗ = ΔΗ = ΔE ΔE + Δ (PV) + nr ΔΗ = nc v + nr ΔΗ = n(c v + R) ΔΗ = nc p

Summary q = nc ΔE = nc v ΔΗ = nc p C v = 3 2 R C p = C v + R

Example: Heating an Ideal Gas Consider 2.00 mol of a monoatomic ideal gas that is taken from state A (P A = 2atm, V A = 10L )to state B (P B = 1atm, V B = 30L) by two different pathways: calculate q, w, ΔE, and ΔH for the two pathways (P A = 2atm, V A = 10L ) P(atm) (P D = 1atm, V D = 10L) Path 1 Path 2 V(L) (P C = 2atm, V C = 30L) (P B = 1atm, V B = 30L)

V A = 10L V C = 30L = nc p = nc v Step 1 At constant pressure = 2atm PΔV = nr PΔV = (2atm) (20L) = 40 atm L x nr = 4.05 x 10 3 J 4.05 x 10 3 J = nr 101.3 J = 4.05 x 10 3 J 1atmL w = -PΔV = -4.05 x 10 3 J ΔH = q p = 1.01 x 10 4 J q p ΔE = n 5 4.05 x 10 3 J R 2 nr = n 3 4.05 x 10 3 J R 2 nr ( )( ) = 1.01 x 104 J ( )( ) A B = 6.08 x 10 3 J

P C = 2atm P B = 1atm At constant volume = 30L ΔPV = nr = (30L)(1atm - 2atm) = = nr w = 0 (no change in volume) q v ΔE = nc v = q v ΔH = nc p Step 2 ΔPV nr -30 atml x nr 3-3.04 x 10 3 J = n R 2 nr A B ( )( ) = -4.56 x 103 J 5-3.04 x 10 3 J = n R 2 nr ( )( ) 101.3 J -3.04 x 10 3 J = 1atmL nr = -7.6 x 10 3 J

P C = 2atm P B = 1atm At constant volume = 10L ΔPV = nr = (10L)(1atm - 2atm) = = nr w = 0 (no change in volume) q v ΔE = nc v = q v ΔH = nc p Step 3 ΔPV nr -10 atml x nr 3-1.01 x 10 3 J = n R 2 nr ( )( ) = -1.52 x 103 J 5-1.01 x 10 3 J = n R 2 nr ( )( ) A B 101.3 J -1.01 x 10 3 J = 1atmL nr = -5.08 x 10 3 J

V A = 10L V C = 30L Step 4 At constant pressure = 1atm PΔV = nr PΔV = (1atm) (20L) = 20 atm L x nr = 2.03 x 10 3 J 2.03 x 10 3 J = nr A B 101.3 J = 2.03 x 10 3 J 1atmL w = -PΔV = -2.03 x 10 3 J ΔH = q p = 5.08 x 10 3 J q p ΔE = nc p = nc v = n 5 2.03 x 10 3 J R 2 nr = n 3 2.03 x 10 3 J R 2 nr ( )( ) = 5.08 x 103 J ( )( ) = 6.08 x 10 3 J

Pathway One A B w = w 1 + w 2 = -4.05 x 10 3 J + 0 q = q 1 + q 2 = -4.56 x 10 3 J + 1.01 x 10 4 J = 5.5 x 10 3 J ΔE = q + w = 5.50 x 10 3 J - 4.05 x 10 3 J = 1.50 x 10 3 J ΔH = ΔH 1 + ΔH 2 = 1.01 x 10 4 J + 7.6 x 10 3 J = 2.50 x 10 3 J

Pathway Two A B w = q = w 3 + w 4 = 0-2.03 x 10 3 J q 3 + q 4 = -1.52 x 10 3 J + 5.08 x 10 4 J = 3.56 x 10 3 J ΔE = q + w = 3.56 x 10 3 J - 2.03 x 10 3 J = 1.5 x 10 3 J ΔH = ΔH 3 + ΔH 4 = 2.53 x 10 3 J + 5.08 x 10 3 J = 2.55 x 10 3 J

w = -4.05 x 10 3 J q = 5.5 x 10 3 J ΔE ΔH = 1.50 x 10 3 J = 2.50 x 10 3 J (P A = 2atm, V A = 10L ) Path 1 P(atm) Path 2 w = q = - 2.03 x 10 3 J 3.56 x 10 3 J V(L) (P B = 1atm, V B = 30L) ΔE = 1.5 x 10 3 J ΔH = 2.55 x 10 3 J