17 1 Model Assume the gas is ideal The work done on a gas is the negative of the area under the pV curve Visualize The gas is compressing so we expect the work to be positive Solve The work done on the gas is W p dV area under the 200 cm 200 kPa pV 3 curve 3 200 10 m 2 0 10 Pa 6 5 40 J Assess The area under the curve is negative because the integration direction is to the left Thus the environment does positive work on the gas to compress it 17 2 Model Assume the gas is ideal The work done on a gas is the negative of the area under the pV curve Visualize The gas is expanding so we expect the work to be negative Solve The area under the pV curve is the area of the rectangle and triangle We have 3 200 10 m 200 10 Pa 6 3 3 200 10 m 200 10 Pa 6 3 1 2 60 J Thus the work done on the gas is W Assess The environment does negative work on the gas as it expands 60 J 17 3 Model Assume the gas is ideal The work done on a gas is the negative of the area under the pV curve Solve The work done on gas in an isobaric process is W p V p V V f i Substituting into this equation 80 J 200 10 Pa 3 3V 1 V 1 V i 3 2 0 10 m 200 cm 4 3 Assess The work done to compress a gas is positive 17 4 Model Helium is an ideal gas that undergoes isobaric and isothermal processes Solve is constant the work done is a Since the pressure p p i p f W on gas p V p V V f i V V f i nRT i V i 0 10 mol 8 31 J mol K 573 K 3 1000 cm 2000 cm 3 2000 cm 3 240 J b For compression at a constant temperature W on gas nRT ln V V f i 0 10 mol 8 31 J mol K 573 K ln 6 1000 10 m 2000 10 m 6 3 3 330 J c For the isobaric case For the isothermal case p i 2 38 10 Pa and the final pressure is 5 p nRT i V i 2 38 10 Pa 5 p f nRT f V f 4 76 10 Pa 5 17 5 Visualize Solve Because W than the initial point so T f E E the gas increases th f p dV and this is an isochoric process W 0 J The final point is on a higher isotherm Heat energy is thus transferred into the gas Q 0 and the thermal energy of i T as the temperature increases th i 17 6 Visualize Solve Because this is an isobaric process W pdV p V V f i Since Vf is smaller than Vi W is positive In other That is the gas is compressed Since the final point is on a lower isotherm than the initial point words the thermal energy decreases For this to happen the heat energy transferred out of the gas must be larger than the work done T i T f 17 7 Visualize Solve Because the process is isothermal E th E th f E 0 J According to the first law of E W Q thermodynamics compressing hence Q is negative That is heat energy is removed from the gas This can only be satisfied if Q th W is positive because the gas is th i W 17 8 Visualize Solve This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the According to the first law of thermodynamics the work done on a gas in an environment That is adiabatic process goes entirely to changing the thermal energy of the gas The work W is positive because the gas is compressed Q 0 J 17 9 Solve The first law of thermodynamics is E W Q 200 J 500 J Q Q Q 700 J th The negative sign means a transfer of energy from the system to the environment Assess Because magnitude than W so that E th i E th f 0W means a transfer of energy into the system Q must be less than zero and larger in 17 10 Solve This is an isobaric process the system Also 100 J of heat energy is transferred out of the gas The first law of thermodynamics is 0W because the gas is compressed This transfers energy into E W Q p V Q 4 0 10 Pa 200 600 10 m 100 J 60 J 6 3 5 th Thermal energy increases by 60 J 17 11 Model The removal of heat from the ice reduces its thermal energy and its temperature Solve The heat needed to change an object s temperature is Q Mc T The mass of the ice cube is M V ice 920 kg m 0 06 0 06 0 06 m 0 1987 kg 3 3 The specific heat of ice from Table 17 2 is c ice 2090 J kg K so Q 0 199 kg 2090 J kg K 243 K 273 K 12 460 J 12 000 J Thus the energy removed from the ice block is 12 000 J Assess The negative sign with Q means loss of energy 17 12 Model The spinning paddle wheel does work and changes the water s thermal energy and its temperature Solve a The temperature change is The mass of the water is 25 C 21 C 4 K T T f T i M 200 10 m 1000 kg m 0 20 kg 6 3 3 W E Mc th water T 0 20 kg 4190 J kg K 4 K 3350 J 3400 J No energy is transferred between the system and the environment because of a difference in The work done is 0 b Q temperature 17 13 Model Heating the mercury at its boiling point changes its thermal energy without a change in temperature Solve The mass of the mercury is c the specific heat mercury 140 J kg K 2 0 10 kg the boiling 20 g M 2 T b 357 C point to the vapor phase is the sum of two steps The first step is and the heat of vaporization L v 2 96 10 J kg The heat required for the mercury to change 5 Q Mc 1 mercury T 2 0 10 kg 140 J kg K 357 C 20 C 940 J 2 The second step is The total heat needed is 6860 J Q ML V 2 2 0 10 kg 2 96 10 J kg 5920 J 2 5 17 14 Model Heating the mercury changes its thermal energy and its temperature Solve a The heat needed to change the mercury s temperature is Q Mc T T Hg Q Mc Hg 100 J 0 020 …
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