Gas Law Problem: Calculating New Volume of Helium
What is the new volume of helium at STP if the temperature is increased to 425.0 K and the pressure is increased to 3.50 atm?
Given data:
Initial volume = 3.20x10^2 mL
Initial temperature (T1) = 0°C or 273 K
Initial pressure (P1) = 1 atm
Final temperature (T2) = 425.0 K
Final pressure (P2) = 3.50 atm
To find the new volume of helium at STP, we can use the ideal gas formula:
\[ \frac{P1 \times V1}{T1} = \frac{P2 \times V2}{T2} \]
Where:
P is pressure
V is volume
T is temperature
Subindex 1 represents initial data and subindex 2 represents final data.
Given the initial volume of 3.20x10^2 mL at STP, we can calculate the new volume using the ideal gas formula:
\[ V2 = \frac{P1 \times V1 \times T2}{T1 \times P2} \]
First, let's convert the initial volume of helium from milliliters to liters since the final volume should be in liters.
Initial volume: 3.20x10^2 mL = 0.32 L
Substitute the given values into the formula:
\[ V2 = \frac{1 \text{ atm} \times 0.32 \text{ L} \times 425.0K}{273 \text{ K} \times 3.50 \text{ atm}} \]
\[ V2 = 0.14 \text{ L} \]
Therefore, the new volume of the helium sample would be 0.14 liters when the temperature is increased to 425.0 K and the pressure is increased to 3.50 atm.