Following figure shows two process A and B for a gas. If `Delta Q_(A)` and `Delta Q_(B)` are the amount of heat absorbed by the system in two case, and `Delta U_(A)` and `Delta U_(B)` are changes in internal energies, respectively, then :

Let Delta U_(1) and Delta U_(2) be the changes in internal energy of an ideal gas in the processes A and B then

If q is the heat added to the system, w is the work done by the system and Delta U is the change in internal energy, then according to first law of thermodynamics :

A system absorbs 300 cal of heat. The work done by the system is 200 cal. Delta U for the above change is

A system absorbs 300 cal of heat. The work done by the system is 200 cal. Delta U for the above change is

An ideal gas is taken from state A to state B via three different processes as shown in the pressure volume (P-V) diagram. If Q_1 , Q_2 & Q_3 indicates the heat absorbed by the gas and Delta U_1 , Delta U_2 & Delta U_3 indicate change in internal energy in three processes, then

For gaseous reaction, if Delta H is the change in enthalpy and Delta U that in internal energy then :

For which reaction, (DeltaH > DeltaU) ? ( Delta H = change in enthalpy, Delta U = change in internal energy)