A pneumatic cylinder in a hydraulic brake mechanism contains 0.050\text{ m}^3 of an ideal … — JAMB UTME

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

$327^\circ\text{C}$ Answer
B
$54^\circ\text{C}$
C
$600^\circ\text{C}$
D
$1227^\circ\text{C}$

Answer

The final temperature of the gas is

327^\circ\text{C}

First, converting the initial temperature gives

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

. For the isothermal compression, the pressure increases to

P_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

P_3 = 6.00 \times 10^5\text{ Pa}

, the pressure doubles, requiring the absolute temperature to double from

300\text{ K}

600\text{ K}

. Converting

600\text{ K}

back to Celsius yields

600 - 273 = 327^\circ\text{C}

Step-by-Step Solution

Convert the initial temperature from Celsius to Kelvin.

T_1 = 27 + 273 = 300\text{ K}

Gas laws strictly require thermodynamic temperatures expressed in Kelvin.

Apply Boyle's Law (

P_1 V_1 = P_2 V_2

) to determine the intermediate pressure

P_2

after isothermal compression at constant temperature

T_2 = 300\text{ K}

P_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.

Apply Pressure Law (Gay-Lussac's Law) for the second stage, where volume remains constant (

V_3 = V_2 = 0.020\text{ m}^3

) while pressure increases from

P_2 = 3.00 \times 10^5\text{ Pa}

P_3 = 6.00 \times 10^5\text{ Pa}

\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.

Convert the final temperature

T_3

back to degrees Celsius.

t_3 = 600 - 273 = 327^\circ\text{C}

The question explicitly requests the final temperature in degrees Celsius.

Key Concept

Multi-stage gas processes combining Boyle's Law and Pressure Law using the Ideal Gas Equation

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