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Air at 27.0°C and at atmospheric pressure is drawn into a bicycle pump that has ...
May 26, 2024
Air at 27.0°C and at atmospheric pressure is drawn into a bicycle pump that has a cylinder with an inner diameter of2.50 cm and a length of 50.0 cm. The down stroke adiabatically compresses the air, which reaches a gauge pressure of 800 kPa before entering the tire. Determine (a) the volume of the compressed air and (b) the temperature of the compressed air. (c) The pump is made of steel and has an inner wall that is 2.00 mm thick. Assume that 4.00 cm ofthe cylinder’s length is allowed to come to thermal equilibrium with the air. What will be the increase in wall temperature?
Generated Graph
Solution by Steps
step 1
Calculate the initial volume of the air in the pump cylinder using the formula for the volume of a cylinder: V=πr2hV = \pi r^2 h. Here, r=2.5 cm2=1.25 cmr = \frac{2.5 \, \text{cm}}{2} = 1.25 \, \text{cm} and h=50 cmh = 50 \, \text{cm}
step 2
V=π(1.25 cm)2(50 cm)=π(1.5625 cm2)(50 cm)=78.54 cm3V = \pi (1.25 \, \text{cm})^2 (50 \, \text{cm}) = \pi (1.5625 \, \text{cm}^2) (50 \, \text{cm}) = 78.54 \, \text{cm}^3
step 3
Convert the volume to cubic meters: V=78.54 cm3×10−6 m3/cm3=7.854×10−5 m3V = 78.54 \, \text{cm}^3 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 7.854 \times 10^{-5} \, \text{m}^3
step 4
Calculate the final volume of the compressed air using the adiabatic process formula: P1V1γ=P2V2γP_1 V_1^\gamma = P_2 V_2^\gamma. Here, P1=101.3 kPaP_1 = 101.3 \, \text{kPa}, P2=800 kPaP_2 = 800 \, \text{kPa}, and γ=1.4\gamma = 1.4
step 5
V2=V1(P1P2)1/γ=7.854×10−5 m3(101.3800)1/1.4=2.56×10−5 m3V_2 = V_1 \left(\frac{P_1}{P_2}\right)^{1/\gamma} = 7.854 \times 10^{-5} \, \text{m}^3 \left(\frac{101.3}{800}\right)^{1/1.4} = 2.56 \times 10^{-5} \, \text{m}^3
step 6
Calculate the temperature of the compressed air using the adiabatic process formula: T2=T1(V1V2)γ−1T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1}. Here, T1=27+273.15=300.15 KT_1 = 27 + 273.15 = 300.15 \, \text{K}
step 7
T2=300.15 K(7.854×10−52.56×10−5)0.4=300.15 K×2.45=735.37 KT_2 = 300.15 \, \text{K} \left(\frac{7.854 \times 10^{-5}}{2.56 \times 10^{-5}}\right)^{0.4} = 300.15 \, \text{K} \times 2.45 = 735.37 \, \text{K}
step 8
Convert the temperature back to Celsius: T2−273.15=735.37−273.15=462.22 °CT_2 - 273.15 = 735.37 - 273.15 = 462.22 \, \text{°C}
step 9
Calculate the increase in wall temperature using the heat transfer formula: Q=mcΔTQ = mc\Delta T. Here, m=ρVm = \rho V, ρ=7850 kg/m3\rho = 7850 \, \text{kg/m}^3, V=πr2h=π(0.0125 m)2(0.04 m)=1.96×10−5 m3V = \pi r^2 h = \pi (0.0125 \, \text{m})^2 (0.04 \, \text{m}) = 1.96 \times 10^{-5} \, \text{m}^3, and c = 500 \, \text{J/kg·K}
step 10
m=7850 kg/m3×1.96×10−5 m3=0.154 kgm = 7850 \, \text{kg/m}^3 \times 1.96 \times 10^{-5} \, \text{m}^3 = 0.154 \, \text{kg}
step 11
ΔT=Qmc=P2V2−P1V1mc=800×103×2.56×10−5−101.3×103×7.854×10−50.154×500=2.56 K\Delta T = \frac{Q}{mc} = \frac{P_2 V_2 - P_1 V_1}{mc} = \frac{800 \times 10^3 \times 2.56 \times 10^{-5} - 101.3 \times 10^3 \times 7.854 \times 10^{-5}}{0.154 \times 500} = 2.56 \, \text{K}
Answer
(a) The volume of the compressed air is 2.56×10−5 m32.56 \times 10^{-5} \, \text{m}^3. (b) The temperature of the compressed air is 462.22 °C462.22 \, \text{°C}. (c) The increase in wall temperature is 2.56 K2.56 \, \text{K}.
Key Concept
Adiabatic Process
Explanation
In an adiabatic process, there is no heat exchange with the surroundings, and the relationship between pressure, volume, and temperature can be used to determine the final state of the gas.
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