Problem 1: Heat Capacity The heat capacity of an ideal mixture of four gases Cpm

Question

Problem 1: Heat Capacity The heat capacity of an ideal mixture of four gases Cpmisture can be expressed in terms of the heat capacity of the components by the mixture equation where x1,32, x3, and x4 are the decimal percentage of the components, and Cp1, Cp2, Cp3, and Cps are the corresponding heat capacities for each of the individual components. A mixture of unknown quantities of the gases S02, SO3, 02, and N2 has been tested by measuring the mixture heat capacity at different temperatures, as provided in the table. Use the data at the three temperatures to create three mathematical relationships base on the mixture equation. In addition, we know that all of the decimal percentages need to sum to be 1.00 (100%), thus the fourth equation in the system is: X1+X2+X3 +X4 = 1 Use the information provided for the respective components and in the description to determine the composition of the mixture of gases, x1, x2, x3, and x4. Heat Capacity Joules/(g mol) °C Mixture so so 25 39.82 39.8667 50.7441 29.3857 29.0514 Temperature [deg C 150 44.72 44.0964 60.4698 30.7047 29.1914 300 49.10 48.0599 69.2528 32.0626 29.0671

Explanation / Answer

Assuming the decimal percentage of SO2 as x1 ; SO3 as x2 ; O2 as x3 ; N2 as x4

Using the table we can form the four equations

x1 + x2 + x3 + x4 = 1 ...............(1)

For temperture 25 degC we can write following equation

39.8667 x1 + 50.7441 x2 + 29.3857 x3 + 29.0514 x4 = 39.82 ...................(2)

For temperture 150 degC we can write following equation

44.0964 x1 + 60.4698 x2 + 30.7047 x3 + 29.1914 x4 = 44.72 .......................(3)

For temperture 300 degC we can write following equation

48.0599 x1 + 69.2528 x2 + 32.0626 x3 + 29.0671 x4 = 49.10 ..............................(4)

Solving the four equation for the four variables we get

x1 = 0.1476

x2 = 0.4212

x3 = 0.1001

x4 = 0.3309

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