A given mass of gas occupies 2dm 3 at 300K. At what temperature will its
A given mass of gas occupies 2dm3 at 300K. At what temperature will its volume be doubled, keeping the pressure constant
- A) 400K
- B) 480K
- C) 550K
- D) 600K
Correct Answer: D) 600K
Explanation
When dealing with gases, we often use the Gas Laws to relate the variables of pressure, volume, and temperature. In this case, we are given the initial volume, temperature, and asked to find the temperature at which the volume is doubled while keeping the pressure constant. The most suitable gas law to use here is Charles's Law.
Charles's Law states that for a given amount of gas with a constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin (K). Mathematically, this law is represented as:
\[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]
Where \(V_1\) and \(T_1\) are the initial volume and temperature, and \(V_2\) and \(T_2\) are the final volume and temperature, respectively.
In this question, we are given:
- \(V_1 = 2 \, dm^3\)
- \(T_1 = 300K\)
- \(V_2 = 2V_1 = 4 \, dm^3\)
We are asked to find the final temperature, \(T_2\). Using Charles's Law, we can plug in the values and solve for \(T_2\):
\[\frac{2 \, dm^3}{300K} = \frac{4 \, dm^3}{T_2}\]Cross-multiplying and solving for \(T_2\), we get:
\[T_2 = \frac{4 \, dm^3 \times 300K}{2 \, dm^3} = 600K\]So the correct answer isOption D: 600K.