Respuesta :
The given question is incomplete. The complete question is as follows.
In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 26.0 ∘C . You identified that the ideal pressure (Pideal) was 49.1 atm , and the real pressure (Preal) was 47.1 atm under these conditions. Complete the sentences to analyze this difference.
The percent difference between the ideal and real gas is _____ .
The difference is considered significant, and is best explained because argon atoms have relatively ______ .
If the volume of container were increased to 2.00 L, you would expect the percent difference between the ideal and real gas to ______ .
Explanation:
The given data is as follows.
Ideal pressure (true value) = 49.1 atm
Real pressure (measured value) = 47.1 atm
Therefore, formula to calculate percent difference is as follows.
Percent difference = [tex]\frac{\text{True value - Measured value}}{\text{True value}} \times 100[/tex]
= [tex]\frac{(49.1 - 47.1) atm}{49.1 atm} \times 100[/tex]
= 4.07%
Thus, we can conclude that the percent difference between the ideal and real gas is 4.07%.
As pressure is inversely proportional to volume according to the ideal gas law. Since, the pressure is low for given argon gas.
Hence, the difference is considered significant, and is best explained because argon atoms have relatively large molecular volumes.
Also, with increase in volume of the container there will be a decrease in pressure of the argon gas. Therefore, probability of interaction between argon atoms decreases.
In this condition, the argon gas tends towards more ideal. Thus, the percent difference between tween the ideal and real gas decreases.
