What is the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm?
PV = nRT
What is the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm?
To find the volume of an ideal gas, you can use the Ideal Gas Law, which is given by the equation:
PV = nRT
Where:
P is the pressure of the gas (in atmospheres)
V is the volume of the gas (in liters)
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas (in Kelvin)
From the question, we are given:
n = 2.0 moles
T = 290 K
P = 2.8 atm
R = 0.0821 L·atm/(mol·K)
We can rearrange the equation to solve for V:
V = (nRT) / P
Now, let's substitute the given values into the equation:
V = (2.0 moles * 0.0821 L·atm/(mol·K) * 290 K) / 2.8 atm
Calculate the expression:
V = (47.8 L·atm) / 2.8 atm
V ≈ 17.07 liters
Therefore, the volume of 2.0 moles of an ideal gas at 290 K and 2.8 atm is approximately 17.07 liters.