A fixed mass of gas in a rigid container exerts a pressure of 3.0\text{ atm} at a temperat… — JAMB UTME

A fixed mass of gas in a rigid container exerts a pressure of $3.0\text{ atm}$ at a temperature of $27^\circ\text{C}$ . If the container is heated at constant volume until the temperature reaches $177^\circ\text{C}$ , what is the final pressure exerted by the gas?

A
$2.0\text{ atm}$
$4.5\text{ atm}$ Answer
C
$19.7\text{ atm}$
D
$1.77\text{ atm}$

Answer

The final pressure exerted by the gas is

4.5\text{ atm}

According to the Pressure Law (Gay-Lussac's Law), the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin gives

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

and

T_2 = 177 + 273 = 450\text{ K}

. Solving for

P_2 = 3.0 \times (450 / 300) = 4.5\text{ atm}

Step-by-Step Solution

Convert all given temperatures from Celsius to Kelvin

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

T_2 = 177 + 273 = 450\text{ K}

Gas laws require absolute temperature measured in Kelvin.

State the Pressure Law equation relating pressure and absolute temperature at constant volume

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

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

Substitute the known values into the equation and solve for final pressure (

P_2

)

P_2 = P_1 \times \frac{T_2}{T_1} = 3.0 \times \frac{450}{300} = 4.5\text{ atm}

Multiplying initial pressure by the ratio of absolute temperatures gives the final pressure.

Key Concept

Pressure Law (Gay-Lussac's Law)

Estimated Time:50s

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