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標題:

Gases

發問:

A mixture of 0.2mol of hydrogen and 0.1mol of oxygen is held in a container which is at a pressure of 1*10^5Nm^-2. The gases are ignited by an electric spark and react to form 0.2 mol of water(in gaseous state). The energy released in the reaction is 300kJmol^-1of water. The initial temperature of the mixture is... 顯示更多 A mixture of 0.2mol of hydrogen and 0.1mol of oxygen is held in a container which is at a pressure of 1*10^5Nm^-2. The gases are ignited by an electric spark and react to form 0.2 mol of water(in gaseous state). The energy released in the reaction is 300kJmol^-1of water. The initial temperature of the mixture is 300K and the molar heat capacity at constant volume for water in the gaseous state may be taken to be 30J mol^-1. The gases are assumed to obey the ideal gas law. Assuming no heating or expansion of the container, what would be the temperature and pressure in the container after the reaction? Q=ms*delta T (300*1000*0.2)=0.2*30*(T-300) T=10300K 但係如果我用 U=3/2nRT 呢條式計, 60000=3/2*0.2*8.31*T T=24067K WHY??? plz help.

最佳解答:

The formula U = (3/2)nRT only applies to monoatomic gas molecules, i.e. molecule with single atom. Such molecules have three degrees of freedom, correspond to the three translational motions. Water vapour molecules are NOT monoatomic, thus the formula cannot be applied. In fact, water vapour molecules have seven degrees of freedom, namely three translational motions, three rotational motions and a vibrational motion of the constituent atoms. Therefore, the internal energy of one mole of water molecules u is u = (7/2).RT and for n moles, the total internal energy U is given by U = (7/2).nRT Turning back to your problem, the energy given to the water vapour = 300 000 x 0.2 J = 60 000 J Thus, 60 000 = (7/2)(0.2)(8.31)(T) T = 10 315 K which matches with the result obtained using specific heat capacity at constant volume.

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