A sample of gas in a flexible container occupies a volume of 4.00\text{ dm}^3 at 27^\circ\… — JAMB UTME

A sample of gas in a flexible container occupies a volume of $4.00\text{ dm}^3$ at $27^\circ\text{C}$ and a pressure of $1.50\text{ atm}$ . What will be the volume of the gas if the temperature is increased to $127^\circ\text{C}$ and the pressure is reduced to $1.00\text{ atm}$ ?

A
$28.2\text{ dm}^3$
$8.00\text{ dm}^3$ Answer
C
$4.50\text{ dm}^3$
D
$6.00\text{ dm}^3$

Answer

The final volume of the gas is

8.00\text{ dm}^3

Applying the General Gas Law

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

requires converting all temperatures to Kelvin:

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

and

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

. Rearranging for

V_2

gives

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1.50 \times 4.00 \times 400}{1.00 \times 300} = 8.00\text{ dm}^3

. Thus, the option stating

8.00\text{ dm}^3

is correct.

Step-by-Step Solution

Convert temperatures from degrees Celsius to Kelvin

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

and

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

Gas laws require absolute temperature units in Kelvin for linear proportional relationships.

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

V_2

)

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

The General Gas Law relates pressure, volume, and temperature for a fixed amount of gas.

Substitute the given values into the rearranged equation

V_2 = \frac{1.50\text{ atm} \times 4.00\text{ dm}^3 \times 400\text{ K}}{1.00\text{ atm} \times 300\text{ K}}

Plugging in

P_1 = 1.50\text{ atm}

V_1 = 4.00\text{ dm}^3

T_1 = 300\text{ K}

P_2 = 1.00\text{ atm}

, and

T_2 = 400\text{ K}

Evaluate the mathematical calculation

V_2 = \frac{2400}{300} = 8.00\text{ dm}^3

Simplifying the fraction gives the final gas volume.

Key Concept

General Gas Law (

P_1 V_1 / T_1 = P_2 V_2 / T_2

)

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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