A sealed cylinder in an industrial pneumatic lift contains 0.080\text{ m}^3 of gas at an i… — JAMB UTME

A sealed cylinder in an industrial pneumatic lift contains $0.080\text{ m}^3$ of gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The piston compresses the gas to a final volume of $0.020\text{ m}^3$ , raising the pressure to $7.50 \times 10^5\text{ Pa}$ . What is the final temperature of the gas in degrees Celsius ( $^\circ\text{C}$ )?

Answer: 102 °C

Answer

The final temperature of the gas is

102^\circ\text{C}

Using the combined gas law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

with absolute initial temperature

T_1 = 300\text{ K}

gives

T_2 = 375\text{ K}

. Subtracting

273

yields the final temperature of

102^\circ\text{C}

Step-by-Step Solution

Convert initial temperature to absolute temperature (Kelvin)

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

Gas laws require thermodynamic temperature measured on the Kelvin scale.

Set up the Combined Gas Law equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Relates initial and final states when pressure, volume, and temperature all change for a fixed mass of gas.

Calculate the final absolute temperature

T_2

T_2 = 300 \times \frac{7.50 \times 10^5 \times 0.020}{1.50 \times 10^5 \times 0.080} = 375\text{ K}

Evaluating the ratio of

P_2 V_2

P_1 V_1

yields

1.25

Convert final temperature from Kelvin to Celsius

\theta_2 = 375 - 273 = 102^\circ\text{C}

The question explicitly requires the answer in degrees Celsius.

Key Concept

Combined Gas Law and thermodynamic temperature conversion

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