Estimate C P , C V , and the difference C P - C V in (J/mol-K)
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Estimate CP, CV, and the difference CP - CV in (J/mol-K) for liquid n-butane from the following data.
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T(°F) 20 40 0 P(psia) 14.7 1400 14.7 V(ft³/1b) 0.02661 0.02662 0.02618 H (BTU/lb) -780.22 -765.05 -791.24 U(BTU/1b) -780.2924302 -771.9507097 -791.3112598
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To estimate the heat capacities at constant pressure CP and at constant volume CV for liquid nbutane we would typically use the following relationships CP dHdTP CV dUdTV where H is enthalpy U is internal energy and T is temperature This requires derivatives of H and U with respect to temperature at constant pressure or volume respectively Unfortunately the data provided doesnt have multiple enthalpy H or internal energy U values at the same pressure or volume to calculate these derivatives directly Since we cant make a direct calculation without additional data or assumptions we have to estimate based on the data at hand We do see data at two different temperatures for the same pressure 147 psia which can be used to crudely estimate CP Assuming the change in enthalpy H is proportional to the change in temperature T we can make the following approximation CP HT where H and T must be in consistent units We have to convert temperature to Kelvin or degrees Celsius for the calculation as entropy and enthalpy changes are typically measured with these units First convert the temperatures from Fahrenheit to Kelvin K TK TF 32 59 27315 T1K 20F 32 59 27315 266483 K T2K 0F 32 59 27315 255372 K Now find the enthalpy change H in the consistent units of Jmol Note 1 BTUlb is approximately 2326 Jg H1 78022 BTUlb H2 79124 BTUlb To convert BTU to J and then from per pound to per mole we need the molecular weight of nbutane which is approximately 5812 gmol H H2 H1 2326 Jg 453592 glb 1 mol5812 g H 79124 78022 2326 453592 5812 Jmol H 1102 2326 453592 5812 Jmol H 1813502 Jmol Now find the temperature change T in Kelvin T T1 T2 266483 K 255372 K 11111 K Now we can estimate CP CP HT 1813502 Jmol 11111 K 163345 JmolK To estimate CV we would need similar data at constant volume which is not provided in the table The difference between CP and CV for an ideal gas is given by the gas constant R which is approximately 8314 JmolK However as were dealing with a liquid and not an ideal gas this relationship ...View the full answer
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Related Book For
Introductory Chemical Engineering Thermodynamics
ISBN: 9780136068549
2nd Edition
Authors: J. Elliott, Carl Lira
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