Chemical Engineering Thermodynamics For an exothremic reaction Only internal energy change (ΔE) is negative Enthalpy change is zero Both ΔH and ΔE are negative Only enthalpy change (ΔH) is negative Only internal energy change (ΔE) is negative Enthalpy change is zero Both ΔH and ΔE are negative Only enthalpy change (ΔH) is negative ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics The expression for entropy change, ΔS = n Cp . ln (T₂/T₁), is valid for the __________ of a substance. Heating Both B and C Cooling Simultaneous pressure & temperature change Heating Both B and C Cooling Simultaneous pressure & temperature change ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics Ideal gas law is applicable at Low T, low P High T, low P High T, high P Low T, high P Low T, low P High T, low P High T, high P Low T, high P ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics Variation of equilibrium pressure with temperature for any two phases of a given substances is given by the __________ equation. Clayperon Gibbs-Duhem Maxwell's None of these Clayperon Gibbs-Duhem Maxwell's None of these ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics A cylinder contains 640 gm of liquid oxygen. The volume occupied (in litres) by the oxygen, when it is released and brought to standard conditions (0°C, 760 mm Hg) will be __________ litres. 22.4 448 224 Data insufficient; can't be computed 22.4 448 224 Data insufficient; can't be computed ANSWER DOWNLOAD EXAMIANS APP
Chemical Engineering Thermodynamics On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line (∂P/∂V)S and the slope of the reversible isothermal line (∂P/∂V)T are related as (where, y = Cp/Cv) ) (∂P/∂V)S = [(∂P/∂V)T]Y (∂P/∂V)S = 1/y(∂P/∂V)T (∂P/∂V)S = y(∂P/∂V)T (∂P/∂V)S = (∂P/∂V)T (∂P/∂V)S = [(∂P/∂V)T]Y (∂P/∂V)S = 1/y(∂P/∂V)T (∂P/∂V)S = y(∂P/∂V)T (∂P/∂V)S = (∂P/∂V)T ANSWER DOWNLOAD EXAMIANS APP