A high-pressure diving cylinder contains 5.00\text{ dm}^3 of compressed air at a pressure … — JAMB UTME

A high-pressure diving cylinder contains $5.00\text{ dm}^3$ of compressed air at a pressure of $3.00\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the air is expanded into a flexible chamber where the pressure drops to $1.50\text{ atm}$ and the temperature rises to $54^\circ\text{C}$ , what is the final volume of the gas?

A
$20.00\text{ dm}^3$
$10.90\text{ dm}^3$ Answer
C
$9.17\text{ dm}^3$
D
$2.73\text{ dm}^3$

Answer

The final volume of the gas is

10.90\text{ dm}^3

Applying the General Gas Law formula

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

with temperatures converted to Kelvin (

T_1 = 300\text{ K}

T_2 = 327\text{ K}

) correctly yields

V_2 = \frac{3.00 \times 5.00 \times 327}{1.50 \times 300} = 10.90\text{ dm}^3

Step-by-Step Solution

Convert initial and final temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 54^\circ\text{C} + 273 = 327\text{ K}

Gas laws strictly require absolute temperature in Kelvin for proportional relationships to hold true.

State the General Gas Law equation and rearrange for the final volume (

V_2

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

The equation relates changes in pressure, volume, and absolute temperature for a fixed mass of gas.

Substitute the given values into the rearranged equation and compute the result.

V_2 = \frac{3.00\text{ atm} \times 5.00\text{ dm}^3 \times 327\text{ K}}{1.50\text{ atm} \times 300\text{ K}} = \frac{4905}{450} = 10.90\text{ dm}^3

Simplifying the numerical expression yields the final expanded volume.

Key Concept

General Gas Law (Combined Gas Law) and Absolute Temperature Scale

Estimated Time:1m 30s

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