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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 _______.

Respuesta :

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 = [tex]\frac{\text{True value - Measured value}}{\text{True value}} \times 100[/tex]

Percent difference = [tex]\frac{(49.3- 47.3)atm}{49.3atm}\times 100[/tex]

Percent difference = 4.06 %

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

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