given 1000 mL of helium at 15degrees C and 763mmHg, determine its volume at -6degrees C and 420 mmHg.
my friend said she got 1.68 x 10^3 mL
help, I have no idea how she got it
She used the ideal gas law, PV = nRT
You can use it in the form PV/T = constant or
V2/V1 = T2/T1 * P1/P2
The temeratures must be absolute (Kelvin)
V2/1000 = 266/288 * 763/420 = 1678 mL
Round it off to 1680
To determine the volume of helium at the given conditions, we can use the combined gas law formula. The combined gas law states:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature (in Kelvin)
P2 = Final pressure
V2 = Final volume (what we are trying to find)
T2 = Final temperature (in Kelvin)
Let's convert the temperatures to Kelvin first:
Initial temperature (T1) = 15 degrees Celsius + 273.15 = 288.15 K
Final temperature (T2) = -6 degrees Celsius + 273.15 = 267.15 K
Now, let's substitute the given values into the formula:
(P1 * V1) / T1 = (P2 * V2) / T2
P1 = 763 mmHg
V1 = 1000 mL
T1 = 288.15 K
P2 = 420 mmHg
V2 = Unknown (what we are trying to find)
T2 = 267.15 K
(763 mmHg * 1000 mL) / 288.15 K = (420 mmHg * V2) / 267.15 K
Now we can solve for V2:
V2 = (763 mmHg * 1000 mL * 267.15 K) / (420 mmHg * 288.15 K)
V2 = 204565450 / 112359.3
V2 ≈ 1820 mL
Therefore, the volume of helium at -6 degrees Celsius and 420 mmHg should be approximately 1820 mL, not 1.68 x 10^3 mL.
To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a gas sample:
(P1 * V1)/ (T1) = (P2 * V2)/(T2)
Here, P1 and P2 are the initial and final pressures respectively, V1 and V2 are the initial and final volumes respectively, and T1 and T2 are the initial and final temperatures respectively.
Let's calculate step by step:
1. Convert the temperatures from Celsius to Kelvin:
Initial temperature (T1) = 15°C + 273.15 = 288.15 K
Final temperature (T2) = -6°C + 273.15 = 267.15 K
2. Convert the pressures from mmHg to atm:
Initial pressure (P1) = 763 mmHg / 760 mmHg/atm = 1.003 atm
Final pressure (P2) = 420 mmHg / 760 mmHg/atm = 0.553 atm
Now we have:
(1.003 atm * V1) / (288.15 K) = (0.553 atm * V2) / (267.15 K)
3. Rearrange the equation to solve for V2:
V2 = (1.003 atm * V1 * 267.15 K) / (0.553 atm * 288.15 K)
Now, substitute the given initial volume (V1 = 1000 mL = 1000 cm^3) into the equation:
V2 = (1.003 atm * 1000 cm^3 * 267.15 K) / (0.553 atm * 288.15 K)
V2 = 1.632 x 10^3 cm^3
Therefore, the volume of helium at -6°C and 420 mmHg is approximately 1.632 x 10^3 mL.
It seems like your friend made a calculation error. The correct answer is approximately 1.632 x 10^3 mL, not 1.68 x 10^3 mL.