A sample of helium has a volume of 521 cc at a pressure of 75 cmHg and a temperature of 18 degrees C. When the temperature is increased to 23 degrees C, what is the volume of the helium?
Vf = .530 L
To answer this question, we can make use of the ideal gas law equation:
PV = nRT
Where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of the gas
- R is the ideal gas constant
- T is the temperature of the gas
In this case, we are given the initial temperature, pressure, and volume of the helium and we need to find the final volume. We can assume that the number of moles of helium remains constant throughout the process.
First, let's convert the initial volume from cc to liters:
Initial volume = 521 cc = 521/1000 L = 0.521 L
Next, we need to convert the pressure from cmHg to atm since the ideal gas constant is in terms of atm:
Initial pressure = 75 cmHg = 75/760 atm
Now we can rearrange the equation to solve for the final volume:
Vf = (Vi * Ti * Tf) / (Pi * Tf)
Substituting the given values:
Vf = (0.521 L * 291 K * 296 K) / (75/760 atm * 291 K)
Vf = (0.521 * 291 * 296) / (75/760 * 291)
Vf ≈ 0.530 L
Therefore, the final volume of the helium when the temperature increases to 23 degrees C is approximately 0.530 L.