help me previous | 5 of 7 I next » can si use ine deal gas law. But at higher pr

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previous | 5 of 7 I next » can si use ine deal gas law. But at higher pressures the deviation from the ideal behavior becomes greater, and we need to take both the size of individual gas particles and the attractive forces between individual gas particles into account. e relationship among its pressure (P), its number of moles ( Significant Figures Feedback: Your answer 4.96 g was either rounded differently or used a different number of significant figures than required for this part. ideal gas law does se properties for s law does not real gases such as olecular volume d at low es. The van der se factors with the ique to each Ideal versus real behavior for gases In the following part you can see how the behavior of real gases deviates from the ideal behavior. You will calculate the pressure values for a gas using the ideal gas law and also the van der Waals equation. Take note of how they differ Part B If 1.00 mol of argon is placed in a 0.500-L container at 25.0 C, what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)? 0.08206 For argon, b 0.03219 L/mol. a 1.345 (L2 atm)/mol and Express your answer to two significant figures and include the appropriate units Hints ] ? P 2.9 atm Submit My Answers Give Up Incorrfect; Try Again; 19 attempts remaining Continue ch 4

Explanation / Answer

We first determine the pressure with the ideal gas law:

PV = nRT

P = nRT / V = (1.0 mol * 0.082 L atm / K mol * (25 + 273.15)K) / 0.5 L = 48.8966 atm

Then we determine pressure with van der waals equation:

[P + (a*(n/V)2)] * (V/n - b) = RT

[P + (1.345 * (1.0/0.5)2)] * (0.5/1.0 - 0.03219) = 0.082 L atm / K mol * (25 + 273.15)K

[P + (1.345 * 4)] * (0.46781) = 24.4483

0.46781P + 2.5168 = 24.4483

P = 46.8812 atm

Finally we calculate the difference:

Pideal - Preal = 48.8966 atm - 46.8812 atm = 2.0154 atm = 2.0 atm

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