Chapter 17

# Chapter 17 - 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

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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 () ( ) 36 3 5 area under the curve 200 cm 200 kPa 200 10 m 2.0 10 Pa 40 J W pdV pV =− =− − = × × = 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.

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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 ( )( ) ( )( ) 63 3 3 1 2 200 10 m 200 10 Pa 60 J −− ×× + = Thus, the work done on the gas is 60 J. W =− Assess: The environment does negative work on the gas as it expands.
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 ( ) fi Wp Vp V V =− Δ =− Substituting into this equation, ( ) () 3 11 80 J 200 10 Pa 3 VV =− × 43 3 i 2.0 10 m 200 cm V ⇒= × = Assess: The work done to compress a gas is positive.

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17.4. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve: (a) Since the pressure if () p pp == is constant the work done is ( ) i on gas f i f i i 33 3 1000 cm 2000 cm 0.10 mol 8.31 J mol K 573 K 240 J 2000 cm nRT Wp V p V V V V V =− Δ =− =− = (b) For compression at a constant temperature, ( ) on gas f i 63 ln 1000 10 m 0.10 mol 8.31 J mol K 573 K ln 330 J 2000 10 m Wn R T V V ⎛⎞ × = ⎜⎟ × ⎝⎠ (c) For the isobaric case, 5 i i 2.38 10 Pa nRT p V × For the isothermal case, 5 i p and the final pressure is 5 f f f 4.76 10 Pa nRT p V ×
17.5. Visualize: Solve: Because Wp d V =− and this is an isochoric process, 0 J. W = The final point is on a higher isotherm than the initial point, so fi . TT > Heat energy is thus transferred into the gas (0 ) Q > and the thermal energy of the gas increases th f th i () EE > as the temperature increases.

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17.6. Visualize: Solve: Because this is an isobaric process () fi W pdV p V V =− . Since V f is smaller than V i , W is positive. That is, the gas is compressed. Since the final point is on a lower isotherm than the initial point, . TT < In other words, the thermal energy decreases. For this to happen, the heat energy transferred out of the gas must be larger than the work done.
17.7. Visualize: Solve: Because the process is isothermal, th th f th i 0J. EE E Δ= − = According to the first law of thermodynamics, th . E WQ Δ=+ This can only be satisfied if . WQ = − W is positive because the gas is compressing, hence Q is negative. That is, heat energy is removed from the gas.

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17.8. Visualize: Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the environment. That is, 0 J. Q = According to the first law of thermodynamics, the work done on a gas in an adiabatic process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas is compressed.
17.9.

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## This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 17 - 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

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