To determine the original pressure of a 750 ml sample of helium (He) at 0 degrees Celsius, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law equation is:
(P1 × V1) / (T1) = (P2 × V2) / (T2)
Where:
P1 and P2 are the initial and final pressures, respectively.
V1 and V2 are the initial and final volumes, respectively.
T1 and T2 are the initial and final temperatures, respectively.
Let's assign the given values:
P1 = unknown (original pressure)
V1 = 750 ml (initial volume)
T1 = 0 degrees Celsius (initial temperature)
P2 = 2 atm (final pressure)
V2 = 500 ml (final volume)
T2 = 25 degrees Celsius (final temperature)
Before using the combined gas law equation, we need to convert the temperatures to Kelvin scale by adding 273.15 to both T1 and T2:
T1 = 0 + 273.15 = 273.15 K
T2 = 25 + 273.15 = 298.15 K
Now we can plug in the values into the combined gas law equation:
(P1 × 750 ml) / (273.15 K) = (2 atm × 500 ml) / (298.15 K)
To solve for P1, we can cross multiply and rearrange the equation:
P1 = (2 atm × 500 ml × 273.15 K) / (750 ml × 298.15 K)
P1 = 0.924 atm
Therefore, the original pressure of the 750 ml sample of helium at 0 degrees Celsius is approximately 0.924 atm.