HonorsHW - Chemistry 443 Honors Question # 1 Fall, 2007...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chemistry 443 Honors Fall, 2007 Question # 1 1-1 DIFFERENCES IN GAS LAWS The ideal-gas law is generally used for gases “if the pressure is sufficiently low”. When this limit is not reached, one has to resort to the so-called real-gas laws, one of which is the gas law of van der Waals. Let’s compare these two laws under different conditions. (a) Calculate the pressure of a gas using the ideal-gas law and the van der Waals-gas law when one mole of carbon dioxide at temperature is 300K is confined in 1 m 3 . If you assume that the van der Waals-gas law gives the “correct” pressure, by what percent is the calculation using the ideal-gas law in error? The pressure is easily calculated by direct substitution for both cases. Ideal gas Torr Pa m K K mol J mol V nRT P 709 . 18 10 49433 . 2 1 300 3144349 . 8 1 3 3 1 1 = × = × × = = This is a low pressure, so one might expect that the van der Waals equation would give a result close to it. van der Waals gas Torr Pa Pa Pa m mol m Pa mol mol m Pa mol m K K mol J mol V a n nb V nRT P 707 . 18 07 . 2494 3658 . 0 10 437 . 2494 ) 1 ( 3658 . 0 ) 1 ( 10 8588 . 42 1 1 300 3144349 . 8 1 3 2 3 2 6 2 1 3 6 3 1 1 2 2 = = × = × × × = = These are extremely close values, so close that one has to carry the calculation to six decimal places. The error in the calculation by using the ideal-gas equation is % 011 . 0 100 707 . 18 707 . 18 709 . 18 % = × = Torr Torr Torr error (b) Repeat the calculations of part (a), except in this case consider that one mole of the gas is confined in a volume of 1 liter (i.e. it is confined in a volume of 0.001 m 3 ). The calculations proceed by the same method here; the numbers are different.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chemistry 443 Honors Fall, 2007 Question # 1 1-2 Ideal gas bar Torr Pa m K K mol J mol V nRT P 9433 . 24 18709 10 49433 . 2 001 . 0 300 3144349 . 8 1 6 3 1 1 = = × = × × = = This pressure is significantly higher than that calculated in part a. The pressure calculated with the van der Waals’s equation might be expected to be different. van der Waals gas bar Pa Pa Pa m mol m Pa mol mol m Pa mol m K K mol J mol V a n nb V nRT P 402 . 22 10 2402 . 2 10 658 . 3 10 60602 . 2 ) 001 . 0 ( 3658 . 0 ) 1 ( 10 8588 . 42 1 001 . 0 300 3144349 . 8 1 6 5 6 2 3 2 6 2 1 3 6 3 1 1 2 2 = × = × × = × × × = = The error made by using the ideal-gas law in this situation is % 3 . 11 100 402 . 22 402 . 22 9433 . 24 % = × = bar bar bar error This is a much larger difference between the two results. (c) Usually, the parameters that we can easily control are temperature and pressure, with the gas volume being the parameter calculated. Calculate the molar volume of CO 2 at 300 K and 1.000 bar, using the ideal-gas law and the van der Waals-gas law. What is the percentage error made by using the ideal-gas law under this condition, if you consider that the van der Waals-gas equation gives the “correct” value?
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 37

HonorsHW - Chemistry 443 Honors Question # 1 Fall, 2007...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online