Physics Solution Manual for 1100 and 2101

# 5 with 5 to determine the final volume vf solving

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Unformatted text preview: 207 × 10 –6 C° –1 W P0 β = = = 4.99 × 10 –6 3 3 Q c ρ 4186 J/ ( kg ⋅ C° ) 1.00 × 10 kg/m ( ) 20. REASONING Since the gas is expanding adiabatically, the work done is given by Equation 15.4 as W = 3 nR (Ti − Tf ) . Once the work is known, we can use the first law of 2 thermodynamics to find the change in the internal energy of the gas. SOLUTION a. The work done by the expanding gas is W = 3 nR (Ti − Tf ) = 2 3 2 ( 5.0 mol ) 8.31 J/ ( mol ⋅ K ) ( 370 K − 290 K ) = +5.0 × 103 J b. Since the process is adiabatic, Q = 0 J, and the change in the internal energy is Chapter 15 Problems 775 ∆U = Q − W = 0 − 5.0 × 103 J = − 5.0 × 103 J 21. SSM REASONING AND SOLUTION a. Since the temperature of the gas is kept constant at all times, the process is isothermal; therefore, the internal energy of an ideal gas does not change and ∆U = 0 . b. From the first law of thermodynamics (Equation 15.1), ∆U = Q − W . But ∆U = 0 , so that Q = W . Since work is done on the gas, the work is negative, and Q = −6.1× 103 J . c. The work done in an isothermal compression is given by Equation 15.3: V W = nRT ln f V i Therefore, the temperature of the gas is T= W ( nR ln Vf / Vi ) = –6.1×103 J = 310 K (3.0 mol)[8.31 J/(mol ⋅ K)] ln (2.5 × 10 –2 m3 )/(5.5 ×10 –2 m3 ) 22. REASONING When n moles of an ideal gas change quasistatically to a final volume Vf from an initial volume Vi at a constant temperature T, the work W done is (see Equation 15.3) V W W = nRT ln f or T= (1) Vi Vf nR ln Vi where R is the universal gas constant. To determine T from Equation (1), we need a value for the work, which we do not have. However, we do have a value for the heat Q. To take advantage of this value, we note that Section 14.3 discusses the fact that the internal energy of an ideal gas is directly proportional to its Kelvin temperature. Since the temperature is constant (the neon expands isothermally), the internal energy remains constant. According to the first law of thermodynamics (Equation 15.1), the change ∆U in the internal energy is given by ∆U = Q – W. Since the internal energy U is constant, ∆U = 0, so that W = Q. SOLUTION Substituting W = Q into the expression for T in Equation (1), we find that the temperature of the gas during the isothermal expansion is 776 THERMODYNAMICS T= Q 4.75 ×103 J = = 208 K Vf 0.250 m3 nR ln ( 3.00 mol ) 8.31 J/ ( mol ⋅ K ) ln 3 0.100 m Vi 23. REASONING We can use the first law of thermodynamics, ∆U = Q − W (Equation 15.1) to find the work W. The heat is Q = −4700 J, where the minus sign denotes that the system (the gas) loses heat. The internal energy U of a monatomic ideal gas is given by U = 3 nRT 2 (Equation 14.7), where n is the number of moles, R is the universal gas constant, and T is the Kelvin temperature. If the temperature remains constant during the process, the internal energy does not change, so ∆U = 0 J. SOLUTION The work done during the isothermal process is W = Q − ∆U = − 4700 J + 0 J = −4700 J The negative sign indicates that work is done on the system. 24. REASONING During an adiabatic process, no heat flows into or out of the gas (Q = 0 J). For an ideas gas, the final pressure and volume (Pf and Vf) are related to the initial pressure and volume (Pi and Vi) by PViγ = Pf Vfγ i ( Equation 15.5 ) , where γ is the ratio of the specific heat capacities at constant pressure and constant volume (γ = 5 3 in this problem). We will use this relation to find Vf /Vi. Solving PViγ = Pf Vfγ for Vf /Vi and noting that the pressure doubles i (Pf /Pi = 2.0) during the compression, we have SOLUTION 1 1 P 1 ( 5 / 3) = i = = 0.66 P Vi f 2.0 Vf γ 25. SSM REASONING When the expansion is isothermal, the work done can be calculated from Equation (15.3): W = nRT ln (Vf / Vi ) . When the expansion is adiabatic, the work done can be calculated from Equation 15.4: W = 3 nR (Ti − Tf ) . 2 Since the gas does the same amount of work whether it expands adiabatically or isothermally, we can equate the right hand sides of these two equations. We also note that since the initial temperature is the same for both cases, the temperature T in the isothermal Chapter 15 Problems 777 expansion is the same as the initial temperature Ti for the adiabatic expansion. We then have V 3 V 3 (T − T ) nRTi ln f = nR (Ti − Tf ) or ln f = 2 i f V 2 V Ti i i SOLUTION Solving for the ratio of the volumes gives Vf = Vi e 3 2 (Ti − Tf ) / Ti = e 2 (405 K − 245 K)/(405 K) = 1.81 3 26. REASONING According to the first law of thermodynamics ∆U = Q − W (Equation 15.1), where ∆U is the change in the internal energy, Q is the heat, and W is the work. This expression may be solved for the heat. ∆U can be evaluated by remembering that the internal energy of a monatomic ideal gas is U = 3 nRT (Equation 14.7), where n is the 2 number of moles, R = 8.31 J/(mol⋅K) is the universal gas constant, and T is the Kelvin temperature. Since heat is being added isothermal...
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