Question

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 1

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 = $\frac{\text{True value - Measured value}}{\text{True value}} \times 100$

Percent difference = $\frac{(49.3- 47.3)atm}{49.3atm}\times 100$

Percent difference = 4.06 %

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