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Zorluk: Çok zorGas Laws and the Ideal Gas Equation

A pneumatic cylinder in a hydraulic brake mechanism contains 0.050 m30.050\text{ m}^3 of an ideal gas at an initial pressure of 1.20×105 Pa1.20 \times 10^5\text{ Pa} and a temperature of 27∘C27^\circ\text{C}. The gas is first compressed isothermally to a volume of 0.020 m30.020\text{ m}^3, and then heated at constant volume until its pressure reaches 6.00×105 Pa6.00 \times 10^5\text{ Pa}. What is the final temperature of the gas in degrees Celsius?

  1. 327∘C327^\circ\text{C}Cevap
  2. B
    54∘C54^\circ\text{C}
  3. C
    600∘C600^\circ\text{C}
  4. D
    1227∘C1227^\circ\text{C}

Cevap

The final temperature of the gas is 327∘C327^\circ\text{C}.
First, converting the initial temperature gives T1=27∘C+273=300 KT_1 = 27^\circ\text{C} + 273 = 300\text{ K}. For the isothermal compression, the pressure increases to P2=P1×V1V2=1.20×105×0.0500.020=3.00×105 PaP_2 = P_1 \times \frac{V_1}{V_2} = 1.20 \times 10^5 \times \frac{0.050}{0.020} = 3.00 \times 10^5\text{ Pa}. Next, during the constant volume heating step to P3=6.00×105 PaP_3 = 6.00 \times 10^5\text{ Pa}, the pressure doubles, requiring the absolute temperature to double from 300 K300\text{ K} to 600 K600\text{ K}. Converting 600 K600\text{ K} back to Celsius yields 600−273=327∘C600 - 273 = 327^\circ\text{C}.

Adım Adım Çözüm

1
Convert the initial temperature from Celsius to Kelvin.
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K}.
Gas laws strictly require thermodynamic temperatures expressed in Kelvin.
2
Apply Boyle's Law (P1V1=P2V2P_1 V_1 = P_2 V_2) to determine the intermediate pressure P2P_2 after isothermal compression at constant temperature T2=300 KT_2 = 300\text{ K}.
P2=P1V1V2=(1.20×105 Pa)(0.050 m3)0.020 m3=3.00×105 PaP_2 = \frac{P_1 V_1}{V_2} = \frac{(1.20 \times 10^5\text{ Pa})(0.050\text{ m}^3)}{0.020\text{ m}^3} = 3.00 \times 10^5\text{ Pa}.
During an isothermal process, the product of pressure and volume remains constant.
3
Apply Pressure Law (Gay-Lussac's Law) for the second stage, where volume remains constant (V3=V2=0.020 m3V_3 = V_2 = 0.020\text{ m}^3) while pressure increases from P2=3.00×105 PaP_2 = 3.00 \times 10^5\text{ Pa} to P3=6.00×105 PaP_3 = 6.00 \times 10^5\text{ Pa}.
P2T2=P3T3  ⟹  T3=T2×P3P2=300 K×6.00×105 Pa3.00×105 Pa=600 K\frac{P_2}{T_2} = \frac{P_3}{T_3} \implies T_3 = T_2 \times \frac{P_3}{P_2} = 300\text{ K} \times \frac{6.00 \times 10^5\text{ Pa}}{3.00 \times 10^5\text{ Pa}} = 600\text{ K}.
At constant volume, pressure is directly proportional to absolute temperature.
4
Convert the final temperature T3T_3 back to degrees Celsius.
t3=600−273=327∘Ct_3 = 600 - 273 = 327^\circ\text{C}.
The question explicitly requests the final temperature in degrees Celsius.

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