A fixed mass of helium gas is sealed inside a rigid container at a pressure of 2.0\text{ a… — JAMB UTME

A fixed mass of helium gas is sealed inside a rigid container at a pressure of $2.0\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the gas is heated until its pressure reaches $3.5\text{ atm}$ while maintaining a constant volume, what is its final absolute temperature in Kelvin?

Cevap:The final absolute temperature of the gas is 【525】 K.

Cevap

The final absolute temperature of the gas is 525 K.

According to Gay-Lussac's Pressure Law, pressure and absolute temperature are directly proportional at constant volume (

P_1/T_1 = P_2/T_2

). First, convert

27^\circ\text{C}

to absolute temperature:

27 + 273 = 300\text{ K}

. Then substitute the pressures and initial temperature into the equation to find

T_2

T_2 = (3.5\text{ atm} \times 300\text{ K}) / 2.0\text{ atm} = 525\text{ K}

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Convert the initial temperature from degrees Celsius to the Kelvin scale.

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

All thermodynamic gas law calculations require absolute temperature in Kelvin.

Apply the Pressure Law (Gay-Lussac's Law) equation for constant volume.

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

The pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature.

Substitute the given values into the formula and solve for the final temperature

T_2

T_2 = \frac{P_2 \times T_1}{P_1} = \frac{3.5 \times 300}{2.0} = 525\text{ K}

Multiplying the new pressure by the initial absolute temperature and dividing by the initial pressure yields the final absolute temperature.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

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