Thermodynamics Test

Mathematical expression for First Law Thermodynamics is/are

For isothermal process

For isothermal irreversible process

Expression for the change in internal energy of a system when no heat is absorbed by the system from the surroundings, but work (w) is done on the system is ______. The type of wall the system has is ____.

Expression for the change in internal energy of a system when no work is done on the system, but q amount of heat is taken out from the system and given to the surroundings is _____. The type of wall the system has is ____.

Expression for the change in internal energy of a system when w amount of work is done by the system and q amount of heat is supplied to the system is _____.  The type of the system is ____.

Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres.
How much heat is absorbed?

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1bar and 100°C is \(41 kJ mol^{–1}\). Calculate the internal energy change in \(kJ mol^{–1}\), when 1 mol of water is vapourised at 1 bar pressure and 100°C. Answer in nearest integer.

A swimmer coming out from a pool is covered with a film of water weighing about 18g. How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at 298K in \(kJ mol^{–1}\). \(\Delta _{vap}H^{\Theta}\) for water at 298K= 44.01 \(kJ mol^{–1}\). Answer in nearest integer.

The enthalpies of all elements in their standard states are:

Calculate the heat in \(10^{-3} kJ \) necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 \(J mol^{–1} K^{–1}\).

The ionization enthalpy of \(Na^+\)  formation from Na(g) is 495.8 \(kJ mol^{–1}\), while the electron gain enthalpy of Br is –325.0 \(kJ mol^{–1}\). Given the
lattice enthalpy of NaBr is –728.4 \(kJ mol^{–1}\). The energy for the formation of NaBr ionic solid is (–) _______ × \(10^{–1}\) \(kJ mol^{–1}\).