At a temperature of 18.0°C of 23.0cm^3 of a certain gas occupies has pressure of 103 kPa.
(a) What volume would the gas occupy at a temperature of 0.00°C and pressure of 101kPa?
(b) 1.00 moles of a certain gas occupies a volume of 22.4 litres at 0.00°C and 101325 Pa. Show that the gas constant R = 8.31 J K^-1 mol^-1.
(c) Determine the number of moles of gas in the 23.0cm^3 sample above.
18+273 = 291 K
0 + 273 = 273 K
P V = n R T
103 * 23 = n R *291
n R = 103*23/291
same n R later
101 V = n R 273
V = n R (273/101cm^3)
V = (103*23/291)(273/101) cm^3
22 cm^3 = .0022 Liters = 22*10^-6m^3
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P V = n R T
101325 (22.4 liters)(1m^3/1000Liters) = 1(R)(273)
R = 8.313846154
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(103000)(23cm^3/10^6cm^3/m^3)=
n (8.31)(291)
n = .103*23/(8.31*291) = 9.8*10^-4 mols =.00098 mols
To answer these questions, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
(a) To find the volume at a different temperature and pressure, we can rearrange the equation as follows:
V1/P1 = V2/P2
Where:
V1 = initial volume
P1 = initial pressure
V2 = final volume (what we are trying to find)
P2 = final pressure
Given:
V1 = 23.0 cm^3
P1 = 103 kPa
T1 = 18.0°C (convert to Kelvin by adding 273.15)
First, convert the temperature to Kelvin:
T1 = 18.0 + 273.15 = 291.15 K
Now we have all the known values to use in the equation:
V1/P1 = V2/P2
Substituting the values:
(23.0 cm^3) / (103 kPa) = V2 / (101 kPa)
Rearranging and solving for V2:
V2 = (23.0 cm^3) * (101 kPa) / (103 kPa) = 22.33 cm^3
Therefore, the volume would be approximately 22.33 cm^3.
(b) We need to show that the gas constant R is equal to 8.31 J K^-1 mol^-1. Given the following information:
V = 22.4 L (convert to cubic meters by multiplying by 0.001)
P = 101325 Pa
T = 0.00°C (convert to Kelvin by adding 273.15)
n = 1.00 moles
First, convert the volume to cubic meters:
V = 22.4 L * 0.001 = 0.0224 m^3
Convert the temperature to Kelvin:
T = 0.00 + 273.15 = 273.15 K
Now we can rearrange the ideal gas law equation to solve for the gas constant R:
PV = nRT
R = PV / (nT)
Substituting the values:
R = (101325 Pa) * (0.0224 m^3) / (1.00 moles * 273.15 K)
Simplifying, we get:
R = 8.31 J K^-1 mol^-1
Therefore, we have shown that the gas constant R is equal to 8.31 J K^-1 mol^-1.
(c) To determine the number of moles of gas in the 23.0 cm^3 sample, we need to rearrange the ideal gas law equation to solve for n:
PV = nRT
n = PV / (RT)
Given:
P = 103 kPa
V = 23.0 cm^3
T = 18.0°C (convert to Kelvin by adding 273.15)
R = 8.31 J K^-1 mol^-1
First, convert the volume to cubic meters:
V = 23.0 cm^3 * 0.000001 = 0.000023 m^3
Convert the temperature to Kelvin:
T = 18.0 + 273.15 = 291.15 K
Now we can substitute the values into the equation:
n = (103 kPa) * (0.000023 m^3) / (8.31 J K^-1 mol^-1 * 291.15 K)
Simplifying, we find:
n ≈ 9.48 * 10^-7 moles
Therefore, the number of moles of gas in the 23.0 cm^3 sample is approximately 9.48 * 10^-7 moles.