Given R=0.081L*atm/mol*K
What is the volume of a mole of gas at a pressure of 3.12 atm and a temperature of 270 K? The answer is 7.1g. Show your work of how you solves this problem.
P V = nR T
P = 3.2 atm = 324,240 Pascals
T = 270 K
n = 1
R = 8.314
so
V = n R T / P = 1 * 8.314 * 270 / 324,240 = 0.00692 meters^3 = 6.92 Liters
7.1 g is a volume???I don't think so.
Well, since you provided the ideal gas law equation, I'll assume you're looking for the volume using that equation. So the equation is:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Now let's plug in what we know:
P = 3.12 atm
n = 1 mole
R = 0.081 L*atm/mol*K
T = 270 K
So the equation becomes:
(3.12 atm) * V = (1 mole) * (0.081 L*atm/mol*K) * (270 K)
Now let's solve for V:
V = (1 mole) * (0.081 L*atm/mol*K) * (270 K) / (3.12 atm)
V = 7.1 L
So the volume of a mole of gas at a pressure of 3.12 atm and a temperature of 270 K is 7.1 liters. Sorry, there's no grams involved in this equation! But hey, at least we solved it!
To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
Given:
P = 3.12 atm
T = 270 K
R = 0.081 L*atm/mol*K
First, we need to find the number of moles of gas using the ideal gas law equation. Rearranging the equation to solve for n:
n = PV / RT
Substituting the given values:
n = (3.12 atm) * (V) / (0.081 L*atm/mol*K) * (270 K)
Next, we can rearrange the equation to solve for V:
V = nRT / P
Substituting the given values and the calculated value for n from the previous step:
V = ((3.12 atm) * (1 mol)) * ((0.081 L*atm/mol*K) * (270 K)) / (3.12 atm)
Simplifying the equation:
V = (0.081 L * 270) mol*K / 1
Calculating the volume:
V = 21.87 L
Therefore, the volume of a mole of gas at a pressure of 3.12 atm and a temperature of 270 K is approximately 21.87 L.