Question
1.30 mol sample of krypton gas at a temperature of 17.0 °C is found to occupy a volume of 24.7 liters. The pressure of this gas sample is
mm Hg.
mm Hg.
Answers
GPT 3.5
To find the pressure of the gas sample, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume (24.7 L)
n = number of moles (1.30 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (17.0°C + 273.15 = 290.15 K)
Now we can plug in the values:
P(24.7) = (1.30)(0.0821)(290.15)
P(24.7) = 31.8969
P ≈ 31.8969 / 24.7
P ≈ 1.29 atm
To convert atm to mm Hg, we can use the conversion factor:
1 atm = 760 mm Hg
So, 1.29 atm x 760 mm Hg = 979.4 mm Hg
Therefore, the pressure of the krypton gas sample is approximately 979.4 mm Hg.
PV = nRT
Where:
P = pressure
V = volume (24.7 L)
n = number of moles (1.30 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (17.0°C + 273.15 = 290.15 K)
Now we can plug in the values:
P(24.7) = (1.30)(0.0821)(290.15)
P(24.7) = 31.8969
P ≈ 31.8969 / 24.7
P ≈ 1.29 atm
To convert atm to mm Hg, we can use the conversion factor:
1 atm = 760 mm Hg
So, 1.29 atm x 760 mm Hg = 979.4 mm Hg
Therefore, the pressure of the krypton gas sample is approximately 979.4 mm Hg.
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