calculate the pressure exerted by 1.00 moles of CO2 confined in a volume of 5.00 L at 450. K.

Compare the pressure with that predicted by the ideal gas equation.

What are you studying? I assume you are to use some equation other than the ideal gas equation FIRST, then compare that with PV = nRT

To calculate the pressure exerted by 1.00 moles of carbon dioxide (CO2) confined in a volume of 5.00 L at 450 K, we can use the ideal gas equation:

PV = nRT

where:
- P is the pressure in units of Pascal (Pa)
- V is the volume in units of cubic meter (m^3)
- n is the number of moles of gas
- R is the ideal gas constant, which is equal to 8.314 J/(mol*K)
- T is the temperature in units of Kelvin (K)

First, let's convert the volume from liters to cubic meters:
1 L = 0.001 m^3
So, 5.00 L = 5.00 * 0.001 m^3 = 0.005 m^3

Now, let's calculate the pressure using the ideal gas equation:
P = (nRT) / V

Plug in the values:
P = (1.00 mol * 8.314 J/(mol*K) * 450 K) / 0.005 m^3

Simplifying the equation:
P = 1.00 * 8.314 * 450 / 0.005 Pa

Calculating further:
P = 747,828 Pa

Therefore, the pressure exerted by 1.00 moles of CO2 confined in a volume of 5.00 L at 450 K is 747,828 Pa.

Now, let's compare this pressure with that predicted by the ideal gas equation. The ideal gas equation predicts the pressure assuming ideal gas behavior, which means that the gas molecules have negligible volume and exert no intermolecular forces. In reality, real gases deviate from ideal behavior at high pressures and low temperatures.

By comparing the calculated pressure with the predicted pressure, we can assess the degree of deviation from ideal gas behavior in this specific case. If the calculated pressure is significantly different from the predicted pressure, it suggests that real gas effects are at play.