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581_PartUniversity Physics Solution

# 581_PartUniversity Physics Solution - 19-16 Chapter 19(b...

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19-16 Chapter 19 (b) 1 1 1 1 2 2 TV T V γ γ = 1 1 1 1 1 1 2 2 T h A T h A γ γ γ γ = 1 0.40 1 2 1 2 0.250 m 300.1 K 479.7 K 207 C 0.0774 m h T T h γ = = = = ° (c) 1 2 ( ) V W nC T T = (Eq.19.25) 4 (20.0 mol)(20.8 J/mol K)(300.1 K 479.7 K) 7.47 10 J W = = − × E VALUATE : In an adiabatic compression of an ideal gas the temperature increases. In any compression the work W is negative. 19.60. I DENTIFY : . m V ρ = The density of air is given by . pM RT ρ = For an adiabatic process, 1 1 1 1 2 2 . TV T V γ γ = pV nRT = S ET U P : Using nRT V p = in 1 1 1 1 2 2 TV T V γ γ = gives 1 1 1 1 2 2 . T p T p γ γ = E XECUTE : (a) The pV -diagram is sketched in Figure 19.60. (b) The final temperature is the same as the initial temperature, and the density is proportional to the absolute pressure. The mass needed to fill the cylinder is then 5 3 6 3 3 0 5 air 1.45 10 Pa (1.23 kg/m )(575 10 m ) 1.02 10 kg. 1.01 10 Pa p m p V p × = = × = × × Without the turbocharger or intercooler the mass of air at 15.0 C T = ° and 5 1.01 10 Pa p = × in a cylinder is 4 0 7.07 10 kg. m V ρ = = × The increase in power is proportional to the increase in mass of air in the cylinder; the percentage increase is 3 4 1.02 10 kg 1 0.44 44%. 7.07 10 kg × = = × (c) The temperature after the adiabatic process is ( 1) / 2 2 1 1 . p T T p γ γ = The density becomes (1 ) / 1/ 1 2 2 2 2 0 0 0 2 1 1 1 1 . T p p p p T p p p p γ γ γ ρ ρ ρ ρ ⎞⎛ = = = ⎟⎜ ⎠⎝ The mass of air in the cylinder is 1 1.40 5 3 6 3 4 5 1.45 10 Pa (1.23 kg/m )(575 10 m ) 9.16 10 kg, 1.01 10 Pa m × = × = × × The percentage increase in power is 4 4 9.16 10 kg 1 0.30 30%. 7.07 10 kg × = = × E VALUATE : The turbocharger and intercooler each have an appreciable effect on the engine power. Figure 19.60 19.61. I DENTIFY : In each case calculate either U Δ or Q for the specific type of process and then apply the first law. (a) S ET U P : isothermal ( 0) T Δ = ; U Q W Δ = 300 J W = + For any process of an ideal gas, . V U nC T Δ = Δ E XECUTE : Therefore, for an ideal gas, if 0 T Δ = then 0 U Δ = and 300 J. Q W = = + (b) S ET U P : adiabatic ( 0) Q = ; U Q W Δ = 300 J W = + E XECUTE : 0 Q = says 300 J U W Δ = − = −

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The First Law of Thermodynamics 19-17 (c) S ET U P : isobaric 0 p Δ = Use W to calculate T Δ and then calculate Q . E XECUTE : ; W p T nR T = Δ = Δ / T W nR Δ = p Q nC T = Δ and for a monatomic ideal gas 5 2 p C R = Thus 5 2 (5 /2)( / ) 5 /2 750 J. Q n R T Rn W nR W = Δ = = = + V U nC T Δ = Δ for any ideal gas process and 3 2 . V p C C R R = = Thus 3 /2 450 J U W Δ = = + E VALUATE : 300 J of energy leaves the gas when it performs expansion work. In the isothermal process this energy is replaced by heat flow into the gas and the internal energy remains the same. In the adiabatic process the energy used in doing the work decreases the internal energy. In the isobaric process 750 J of heat energy enters the gas, 300 J leaves as the work done and 450 J remains in the gas as increased internal energy.
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