At a temperature of 27^\circ\text{C}, the root-mean-square (r.m.s.) speed of the molecules… — JAMB UTME

At a temperature of $27^\circ\text{C}$ , the root-mean-square (r.m.s.) speed of the molecules of an ideal gas is $300\text{ m/s}$ . What is the temperature of the gas, in degrees Celsius, when the r.m.s. speed of its molecules increases to $600\text{ m/s}$ ?

Answer: 927 °C

Answer

The temperature of the gas when the r.m.s. speed reaches

600\text{ m/s}

927^\circ\text{C}

According to kinetic theory, the root-mean-square speed of gas molecules is directly proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

). First convert the initial temperature to Kelvin:

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

. Since the speed doubles from

300\text{ m/s}

600\text{ m/s}

, the ratio of speeds is

2

, which means the absolute temperature ratio is

2^2 = 4

. Thus, the new absolute temperature is

4 \times 300\text{ K} = 1200\text{ K}

. Converting back to Celsius gives

1200 - 273 = 927^\circ\text{C}

Step-by-Step Solution

Convert the initial temperature from Celsius to Kelvin

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

Gas kinetic equations require absolute temperature in Kelvin.

Apply the proportional relationship between r.m.s. speed and absolute temperature

\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}

In the kinetic theory of gases, root-mean-square speed is directly proportional to the square root of absolute temperature.

Calculate the final absolute temperature

T_2

T_2 = 1200\text{ K}

Doubling the r.m.s. speed requires quadrupling the absolute temperature (

2^2 \times 300\text{ K} = 1200\text{ K}

Convert the calculated absolute temperature back to degrees Celsius

\theta_2 = 1200 - 273 = 927^\circ\text{C}

Subtract 273 from the Kelvin temperature to find the value in degrees Celsius.

Key Concept

Proportionality between root-mean-square speed and absolute temperature in kinetic theory of gases

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