movable chamber has a volume of 18.5 L (at a temperature of 18.5 C) Assuming no gas escapes and the pressure remains constant; what is the temperature when the chamber has a volume of 19.8 L? Your answer should have 3 significant digits: Do not include units in your answer: Enter the answer’
The Correct Answer and Explanation is:
To solve this, we use Charles’s Law, which states:V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}T1V1=T2V2
Where:
- V1=18.5 LV_1 = 18.5 \, \text{L}V1=18.5L (initial volume)
- T1=18.5∘C=291.65 KT_1 = 18.5^\circ C = 291.65 \, \text{K}T1=18.5∘C=291.65K (initial temperature in Kelvin)
- V2=19.8 LV_2 = 19.8 \, \text{L}V2=19.8L (final volume)
- T2=?T_2 = ?T2=? (final temperature in Kelvin)
Step-by-step solution:
Rewriting Charles’s Law to solve for T2T_2T2:T2=V2⋅T1V1T_2 = \frac{V_2 \cdot T_1}{V_1}T2=V1V2⋅T1T2=19.8⋅291.6518.5T_2 = \frac{19.8 \cdot 291.65}{18.5}T2=18.519.8⋅291.65T2=5774.6718.5≈312.69 KT_2 = \frac{5774.67}{18.5} \approx 312.69 \, \text{K}T2=18.55774.67≈312.69K
Convert back to Celsius:T2∘C=312.69−273.15=39.5T_2^\circ C = 312.69 – 273.15 = \boxed{39.5}T2∘C=312.69−273.15=39.5
✅ Final Answer (3 significant digits, no units):
39.5
📘 Explanation
This problem involves understanding how gases behave when temperature and volume change under constant pressure. Charles’s Law describes this relationship: the volume of a gas is directly proportional to its absolute temperature when pressure is constant. That means, if temperature increases, volume increases — and vice versa — but only if temperature is measured in Kelvin, not Celsius.
First, we convert the given Celsius temperature to Kelvin. This is essential because gas laws are based on absolute temperature scales. We add 273.15 to 18.5°C, giving us 291.65 K.
Next, we apply Charles’s Law:V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}T1V1=T2V2
This formula can be rearranged to find the unknown temperature T2T_2T2, which represents the final temperature when the gas expands to 19.8 L.
By substituting the known values into the equation:T2=19.8×291.6518.5T_2 = \frac{19.8 \times 291.65}{18.5}T2=18.519.8×291.65
We calculate T2T_2T2 to be approximately 312.69 K. We then convert this back to Celsius by subtracting 273.15, yielding 39.5°C.
The significant digits are based on the precision of the input values (3 significant digits), so the final answer must match this precision. Therefore, the correct answer is 39.5, with no units included as instructed.
