The absorption of heat by the system tends to raise the energy of the system. The performance of work by the system, on the other hand, tends to lower the energy of the system because performance of work requires expenditure of energy. Therefore the change ...

As we know that if the quantity of heat transferred from the surrounding to the system is q and work done in the process is w, then the change in internal energy, ΔU = q + w where heat absorbed ...

We can calculate the change in change in thermodynamic properties like q , w , ∆U , ∆H with the help of first law of thermodynamics. The expansion can be isothermal which can be reversible or irreversible or it can be adiabatic which can also ...

Temperature T = 37oC = 37 + 273 = 310 K Since the process is Isothermal, Therefore, ∆U=0 and ∆H = 0 (as for an isothermal expansion of an ideal gas ∆U=0 and ∆H = 0) As work done in reversible ...

(a) We know that for one mole of an ideal gas, CV = (∂U/∂T) V Therefore, dU = CV dT For a finite change, ∆U = CV ∆T As for an isothermal process, T is constant so that ∆T = 0. Hence for an isothermal expansion of an ideal gas ∆U=0. (b) We ...

Relation between CP and CV in gaseous systems No external work is being done when a gas is heated at constant volume i.e. gas uses all the haet which is given to it for increasing its internal energy. Hence if temperature of one mole of a ...

Enthalpy of Vaporizations: During evaporation of liquid, some absorption of heat from surroundings takes place. Hence, Liquid evaporation is fulfilled by increase in enthalpy. The increase in enthalpy while evaporation of one mole of water at 25oC is 43.93 kJ. This result can be expressed in the form ...

We know that work differential dw is calculated as: dw= PdV ……………. (1) since volume V is a state function i.e. it is independent of the path [V=f (T,P)] hence dV is an exact differential.Therefore ……. (2) For an ideal gas, we know ...