A fixed mass of an ideal gas occupies a volume of 3.0\text{ dm}^3 at 27^\circ\text{C} and … — JAMB UTME

A fixed mass of an ideal gas occupies a volume of $3.0\text{ dm}^3$ at $27^\circ\text{C}$ and a pressure of $1.0\text{ atm}$ . What is the final volume of the gas in $\text{dm}^3$ when the temperature is increased to $127^\circ\text{C}$ while maintaining constant pressure?

Answer: 4.0 / 4 / 4.0 dm3 / 4 dm3 / 4.0 dm³ / 4 dm³

Answer

The final volume of the gas is

4.0\text{ dm}^3

First convert temperatures to Kelvin:

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

and

T_2 = 127 + 273 = 400\text{ K}

. At constant pressure,

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

. Rearranging to solve for

V_2

gives

V_2 = \frac{3.0 \times 400}{300} = 4.0\text{ dm}^3

Step-by-Step Solution

Convert both temperatures from degrees Celsius to Kelvin

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

and

T_2 = 127 + 273 = 400\text{ K}

Gas law calculations strictly require absolute temperature measured in Kelvin.

Apply Charles's Law derived from the ideal gas equation (

PV = nRT

) at constant pressure

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

Volume is directly proportional to absolute temperature when pressure and amount of gas are constant.

Substitute the values into the formula and solve for

V_2

V_2 = \frac{V_1 \times T_2}{T_1} = \frac{3.0 \times 400}{300} = 4.0\text{ dm}^3

Calculates the new gas volume.

Key Concept

General Gas Law and Absolute Temperature Conversion

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