Chemical Engineering Thermodynamics Charles' law for gases states that V/T = Constant PV/T = Constant V ∝ 1/T V ∝ 1/P V/T = Constant PV/T = Constant V ∝ 1/T V ∝ 1/P ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics A reasonably general expression for vapour-liquid phase equilibrium at low to moderate pressure is Φi yi P = Yi xi fi° where, Φ is a vapor fugacity component, Yi is the liquid activity co-efficient and fi° is the fugacity of the pure component i. the Ki value (Yi = Ki xi) is therefore, in general a function of Temperature only Temperature, pressure and liquid composition xi only Temperature, pressure, liquid composition xi and vapour composition yi Temperature and pressure only Temperature only Temperature, pressure and liquid composition xi only Temperature, pressure, liquid composition xi and vapour composition yi Temperature and pressure only ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics Law of corresponding states says that No gas can be liquified above the critical temperature, howsoever high the pressure may be The molar heat of energy of gas at constant volume should be nearly constant (about 3 calories) The surface of separation (i. e. the meniscus) between liquid and vapour phase disappears at the critical temperature , two different gases behave similarly, if their reduced properties (i.e. P, V and T) are same No gas can be liquified above the critical temperature, howsoever high the pressure may be The molar heat of energy of gas at constant volume should be nearly constant (about 3 calories) The surface of separation (i. e. the meniscus) between liquid and vapour phase disappears at the critical temperature , two different gases behave similarly, if their reduced properties (i.e. P, V and T) are same ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics The Maxwell relation derived from the differential expression for the Helmholtz free energy (dA) is (∂S/∂V)T = (∂P/∂T)V (∂S/∂P)T = -(∂V/∂T)P (∂T/∂V)S = -(∂P/∂S)V (∂V/∂S)P = (∂T/∂P)S (∂S/∂V)T = (∂P/∂T)V (∂S/∂P)T = -(∂V/∂T)P (∂T/∂V)S = -(∂P/∂S)V (∂V/∂S)P = (∂T/∂P)S ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics During Joule-Thomson expansion of gases Enthalpy remains constant Temperature remains constant None of these Entropy remains constant Enthalpy remains constant Temperature remains constant None of these Entropy remains constant ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics In any spontaneous process, Only A decreases Both F and A decreases Only F decreases Both F and A increase Only A decreases Both F and A decreases Only F decreases Both F and A increase ANSWER DOWNLOAD EXAMIANS APP
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