IN A CERTAIN EXPERIEMT A SAMPLE OF HELIUM IN A VACUUM SYSTEM WAS COMPRESSED AT 25C FROM A VOLUME OF 6000.0ML TO A VOLUME OF 240.0ML WHERE ITS PRESSURE WAS FOUND TO BE 32.0 MM HG. WAHT IS THE ORIGINAL PRESSURE OF HELIUM?
(P1V1)/T1 (P2V2)/T2
Change T to Kelvin.
if the value of kc is lower the product will be unprofitabble
To find the original pressure of helium, we can use the ideal gas law formula:
(P1 * V1) / T1 = (P2 * V2) / T2
Given:
Initial volume (V1) = 6000.0 mL
Final volume (V2) = 240.0 mL
Final pressure (P2) = 32.0 mm Hg
Temperature (T1) = 25°C (convert to Kelvin)
Let's calculate:
Step 1: Convert temperature from Celsius to Kelvin.
To convert temperature from Celsius to Kelvin, we add 273.15 to the Celsius value.
T1 = 25°C + 273.15 = 298.15 K
Step 2: Plug in the values into the formula.
(P1 * V1) / T1 = (P2 * V2) / T2
(P1 * 6000.0 mL) / 298.15 K = (32.0 mm Hg * 240.0 mL) / 298.15 K
Step 3: Solve for P1 (original pressure).
P1 = ((32.0 mm Hg * 240.0 mL) / 298.15 K) * (298.15 K / 6000.0 mL)
P1 = ((32.0 * 240.0)/(298.15 * 6000.0)) * 298.15
P1 ≈ 1.555 mm Hg
Therefore, the original pressure of helium is approximately 1.555 mm Hg.