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.
change temps to Kelvins
P2V2/T2=P1V1/T1
a) solve for V2
b. pv=nRT change V to m^3 (22.4E-3)
R=PV/nT=101325*22.4E-3/(1*273)=...
c. PV=nRT
n=PV/RT
(a) To find the volume the gas would occupy at a temperature of 0.00°C and a pressure of 101 kPa, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 = initial pressure = 103 kPa
V1 = initial volume = 23.0 cm^3
T1 = initial temperature = 18.0°C + 273.15 (converted to Kelvin)
P2 = final pressure = 101 kPa
V2 = final volume (to be determined)
T2 = final temperature = 0.00°C + 273.15 (converted to Kelvin)
Now we can plug in the values and solve for V2:
(103 kPa * 23.0 cm^3) / (18.0°C + 273.15) = (101 kPa * V2) / (0.00°C + 273.15)
Cross multiplying and rearranging the equation, we have:
V2 = [(103 kPa * 23.0 cm^3) * (0.00°C + 273.15)] / (101 kPa * (18.0°C + 273.15))
Simplifying the equation gives us:
V2 = 23.5 cm^3
Therefore, the gas would occupy a volume of 23.5 cm^3 at a temperature of 0.00°C and a pressure of 101 kPa.
(b) To show that the gas constant, R, is equal to 8.31 J K^-1 mol^-1, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure = 101325 Pa
V = volume = 22.4 L
n = number of moles = 1.00 mol
R = gas constant (to be determined)
T = temperature = 0.00°C + 273.15 (converted to Kelvin)
Substituting the values into the equation gives us:
(101,325 Pa * 22.4 L) = (1.00 mol * R * (0.00°C + 273.15))
Simplifying the equation, we have:
R = (101,325 Pa * 22.4 L) / (1.00 mol * (0.00°C + 273.15))
Converting to the proper units:
R = (101,325 N/m^2 * 22.4 m^3) / (1.00 mol * 273.15 K)
Simplifying further, we get:
R = 8.31 J K^-1 mol^-1
Therefore, the gas constant, R, equals 8.31 J K^-1 mol^-1.
(c) To determine the number of moles of gas in the 23.0 cm^3 sample, we can use the ideal gas equation:
PV = nRT
where:
P = pressure = 103 kPa
V = volume = 23.0 cm^3
n = number of moles (to be determined)
R = gas constant = 8.31 J K^-1 mol^-1
T = temperature = 18.0°C + 273.15 (converted to Kelvin)
Converting the volume to liters, we have:
V = 23.0 cm^3 = 0.023 L
Substituting the values into the equation gives us:
(103 kPa * 0.023 L) = n * (8.31 J K^-1 mol^-1) * (18.0°C + 273.15)
Simplifying the equation, we have:
(2.369 kPa*L) = n * (8.31 J K^-1 mol^-1) * (291.15 K)
Rearranging and solving for n, we get:
n = (2.369 kPa*L) / [(8.31 J K^-1 mol^-1) * (291.15 K)]
Converting kPa to Pa and J to kJ in the equation:
n = (2.369 * 10^3 Pa * 0.023 L) / [(8.31 * 10^-3 kJ K^-1 mol^-1) * (291.15 K)]
Simplifying further, we have:
n = 0.023 mol
Therefore, the number of moles of gas in the 23.0 cm^3 sample is 0.023 mol.