In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 27.0 ∘C . You identified that the ideal pressure (Pideal) was 49.3 atm , and the real pressure (Preal) was 47.3 atm under these conditions. The percent difference between the ideal and real gas is .

Answer : The percent difference between the ideal and real gas is, 4.06 %.

Explanation : Given,

Ideal pressure (true value) = 49.3 atm

Real pressure (measured value) = 47.3 atm

The formula used to calculate percent difference is :

Percent difference =

Percent difference =

Percent difference = 4.06 %

Therefore, the percent difference between the ideal and real gas is, 4.06 %.

the answer is d. 74.02

you gotta add the following molecules so we can give u an answer

explanation: