At what tempature will .0100 moles of argon gas have a volume of 275mL at 100 kPa? please this is due tommorrow really need som help
Use PV = nRT. T will be in kelvin. P should be in atmospheres and you have it in kPa. 1 atm = 101.325 kPa. Substitute V in L, not mL.
331 k
To find the temperature at which 0.0100 moles of argon gas has a volume of 275 mL at 100 kPa, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kPa)
V = volume (in L)
n = number of moles
R = ideal gas constant (8.31 J/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given volume from milliliters to liters by dividing it by 1000:
V = 275 mL ÷ 1000 = 0.275 L
Substituting the given values into the ideal gas law equation:
(100 kPa) * (0.275 L) = (0.0100 moles) * (R) * (T)
Rearranging the equation and solving for T:
T = (P * V) / (n * R)
T = (100 kPa * 0.275 L) / (0.0100 moles * 8.31 J/(mol·K))
Calculating the numerical value:
T = (100 kPa * 0.275 L) / (0.0100 moles * 8.31 J/(mol·K))
T = 2750 J / (0.0831 J/(mol·K))
T = 33,113.02 K
Therefore, at 100 kPa, 0.0100 moles of argon gas will have a volume of 275 mL at a temperature of about 33,113.02 Kelvin.
To find the temperature at which 0.0100 moles of argon gas will occupy a volume of 275 mL at 100 kPa, we can use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in kPa)
V = volume (in liters)
n = number of moles
R = ideal gas constant
T = temperature (in Kelvin)
First, we need to convert the given volume to liters. Since 1 L equals 1000 mL, the volume becomes:
V = 275 mL = 275/1000 L = 0.275 L
Next, we can rearrange the Ideal Gas Law equation to solve for T:
T = PV / (nR)
Now, let's substitute the given values into the equation:
P = 100 kPa
V = 0.275 L
n = 0.0100 moles
R = 8.314 J/(mol·K) (ideal gas constant)
Before we proceed, we need to convert the pressure from kPa to atm, as the value of R we have is based on atm:
1 atm = 101.325 kPa (approximately)
P = 100 kPa / 101.325 kPa/atm = 0.9869 atm
Now, we can plug the values into the equation:
T = (0.9869 atm) * (0.275 L) / (0.0100 moles * 8.314 J/(mol·K))
Calculating this, we get:
T ≈ 32.74 K
Therefore, at approximately 32.74 Kelvin, 0.0100 moles of argon gas will occupy a volume of 275 mL at 100 kPa.