Acetylene gas, C2H2 is used for welding. A 5 liter supply of acetylene being stored at -23 °C, exerts a pressure of 5 atm. At what temperature would the same number of moles of acetylene, moved to a 10 liter container, produce a pressure of 2 atm? Make sure to include all steps to your problem! 1 point for showing work, 1 point for the correct final answer.
3 years ago
3 years ago
@bobpursley i dont know the volume
11 months ago
To solve this problem, we can use the ideal gas law, which is expressed as:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
We are given:
P1 = 5 atm
V1 = 5 L
n1 (initial number of moles) = ?
T1 = -23 °C
We need to find:
T2 (final temperature)
First, we convert the initial temperature from Celsius to Kelvin. The Kelvin scale is the absolute temperature scale, where 0 K is the lowest possible temperature (equivalent to -273.15 °C). To convert Celsius to Kelvin, we add 273.15:
T1 = -23 °C + 273.15 = 250.15 K
Next, we can rearrange the equation to solve for n1:
n1 = PV1 / (RT1)
Substituting the given values:
n1 = (5 atm * 5 L) / (0.0821 L·atm/mol·K * 250.15 K)
n1 ≈ 1.009 moles
Now, we can use the same number of moles in the final container to find the final temperature. Rearranging the ideal gas law equation again:
T2 = PV2 / (n1 * R)
Substituting the given values:
T2 = (2 atm * 10 L) / (1.009 moles * 0.0821 L·atm/mol·K)
T2 ≈ 247.2 K
Therefore, the final temperature is approximately 247.2 Kelvin.