Multiple Choice
Which of the following equations represents Boyle's Law?
7. Gases
Chemistry Gas Laws
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What was the original temperature of a gas, in °C, if the volume changed from 1.00 L to 1.10 L and the final temperature was 255.0 °C, assuming constant pressure?
A
231.8 °C
B
245.0 °C
C
260.0 °C
D
275.0 °C
Verified step by step guidance1
Identify the gas law applicable to the problem. Since the pressure is constant and we are dealing with volume and temperature changes, use Charles's Law, which states that \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \), where \( V \) is volume and \( T \) is temperature in Kelvin.
Convert the final temperature from Celsius to Kelvin. The formula for conversion is \( T(K) = T(°C) + 273.15 \). So, the final temperature \( T_2 = 255.0 + 273.15 \).
Substitute the known values into Charles's Law. You have \( V_1 = 1.00 \text{ L} \), \( V_2 = 1.10 \text{ L} \), and \( T_2 \) in Kelvin from the previous step. The equation becomes \( \frac{1.00}{T_1} = \frac{1.10}{T_2} \).
Solve for the initial temperature \( T_1 \) in Kelvin by rearranging the equation: \( T_1 = \frac{1.00 \times T_2}{1.10} \).
Convert the initial temperature \( T_1 \) from Kelvin back to Celsius using the formula \( T(°C) = T(K) - 273.15 \).
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