A fixed amount of gas at 25.0 °C occupies a volume of 10.0 L when the pressure is 629 torr.
Using Charles’s law, calculate the volume (L) the gas will occupy when the temperature is increased to 121 °C while maintaining the pressure at 629 torr.
The Correct Answer and Explanation is :
To solve this problem, we use Charles’s Law, which states that for a fixed amount of gas at constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin. Mathematically, Charles’s Law is given as:
[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
]
where:
- ( V_1 ) and ( V_2 ) are the initial and final volumes of the gas,
- ( T_1 ) and ( T_2 ) are the initial and final temperatures in Kelvin.
Step 1: Convert Temperatures to Kelvin
To use Charles’s Law, we need temperatures in Kelvin:
- Initial temperature, ( T_1 = 25.0^\circ \text{C} + 273.15 = 298.15 \, \text{K} )
- Final temperature, ( T_2 = 121^\circ \text{C} + 273.15 = 394.15 \, \text{K} )
Step 2: Set up Charles’s Law
The initial volume ( V_1 ) is 10.0 L, and we want to find the final volume ( V_2 ) at the higher temperature while maintaining the same pressure (629 torr).
Using Charles’s Law:
[
\frac{10.0 \, \text{L}}{298.15 \, \text{K}} = \frac{V_2}{394.15 \, \text{K}}
]
Step 3: Solve for ( V_2 )
Rearrange to find ( V_2 ):
[
V_2 = \frac{10.0 \, \text{L} \times 394.15 \, \text{K}}{298.15 \, \text{K}}
]
Calculating this:
[
V_2 \approx 13.22 \, \text{L}
]
Explanation
Charles’s Law helps us understand that when the temperature of a gas increases, the kinetic energy of gas particles also increases. This energy increase causes particles to move more rapidly, exerting greater force on the container’s walls, and the gas expands if pressure remains constant. Here, we increased the gas temperature from 25°C to 121°C, causing a corresponding increase in volume from 10.0 L to 13.22 L. By maintaining pressure constant and only changing temperature, Charles’s Law predicts this volume change in a predictable, linear relationship, as we observed with our calculation.