Answer:
The new volume of the [tex] NH_{3}[/tex] sample is 3.27mL.
Explanation:
Given the following data;
Original Volume, V1 = 2.50-mL
Original Temperature, T1 = 325K
New Temperature, T2 = 425K
To find new volume V2, we would use Charles' law.
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
[tex] VT = K[/tex]
[tex] \frac{V1}{T1} = \frac{V2}{T2}[/tex]
Making V2 as the subject formula, we have;
[tex] V_{2}= \frac{V1}{T1} * T_{2}[/tex]
[tex] V_{2}= \frac{2.50}{325} * 425[/tex]
[tex] V_{2}= 0.0077 * 425[/tex]
V2 = 3.27mL
Therefore, new volume of the [tex] NH_{3}[/tex] sample is 3.27mL.
