उष्मागतिकी (Thermodynamics)|तापीय साम्य या तापीय संतुलन और ताप (Thermal Equilibrium)|उष्मागतिकी का शून्यावाँ या शून्य कोटि नियम (Zeroth Law of Thermodynamics)|सूचक आरेख (Indicator Diagram)|प्रसार में किया गया कार्य (Work done During Expansion)|समतापी प्रक्रम (Isothermal Process)|समतापी प्रक्रम में किया गया कार्य (Work done in Isothermal Process)|रुद्धोष्म प्रक्रम (Adiabatic Process)|रुद्धोष्म प्रक्रम मे किया गया कार्य (Work done in Adiabatic Process)|OMR

100 moles of an ideal monatomic gas undergoes the thermodynamic process as shown in the figure ArarrB: isothermal expansion BrarrC: adiabatic expansion CrarrD: isobaric compression DrarrA: isochoric process The heat transfer along the process AB is 9xx10^4J . The net work done by the gas during the cycle is [Take R=8JK^-1mol^-1 ]

The first law of thermodynamics was given as q=DeltaU+(-w) , where q is heat given to a system and DeltaU represents increase in internal energy and -w is work done by the system. Various processes such as isothermal, adiabatic, cyclic, isobaric and isochoric process in terms of I law of thermodynamics leads for important results. The molar heat capacity for 1 mole of monoatomic gas is 3/2 R at constant volume and 5/2 R at constant pressure. A system is allowed to move from state A to B following path ACB by absorbing 80J of heat energy. The work done by the system is 30J . The work done by the system in reaching state B from A is 10J through path ADB, Which statements are correct? (1) Increase in internal energy from state A to state B is 50J . (2) If path ADB is followed to reach state B, DeltaU=50J . (3) If work done by the system in path AB is 20J, the heat absorbed during path AB=70J . (4) The value U_C - U_A is equal to U_D - U_B . (5) Heat absorbed by the system to reach B from A through path ADB is 60J .

The first law thermodynamics was gives as q=Delta U+(-w) , where q is heat given to a system and Delta U represent increase in internal energy and -w is work done by the system. Various process such as isothermal, adiabatic, cyclic, isobaric and isochoric process in terms to first law of thermodynamics leads for important results. The molar heat capacity for 1 mole of monoatomic gas is 3/2R at constant volume and 5/2R at constant pressure. A system is allowed to move from state A to B following path ACB by absorbing 80 J of heat energy. The work done by the system is 30 J. The work done by the system in reaching state B from A is 10 J through path ADB which statements are correct : 1. Increase in internal energy from state A to state B is 50 J. 2. If path ADB is followed to reach state B. Delta U = 50J 3. If work done by the system in path AB is 20 J, the heat absorbed during path AB = 70 J. 4. The value U_(C)-U_(A) is equal to U_(D)-U_(B) . 5. Heat absorbed by the system to reach B from A through path ADB is 60 J.

The pressure-volume of varies thermodynamic process is shown in graphs: Work is the mole of transference of energy. It has been observed that reversible work done by the system is the maximum obtainable work. w_(rev) gt w_(irr) The works of isothermal and adiabatic processes are different from each other. w_("isothermal reversible") = 2.303 nRT log_(10) ((V_(2))/(V_(1))) = 2.303 nRT log_(10)((P_(2))/(P_(1))) w_("adiabatic reversible") = C_(V) (T_(1)-T_(2)) A thermodynamic system goes in a cyclic process as represented in the following P -V diagram: The network done during the complete cycle is given by the area

Work is the mose of transfrence of energy. If the system involves gaseous substance and there is difference of pressure between system and surroundings, such a work is referred to as pressure - volume work (W_(PV)= -P_(ext)DeltaV) . It has been observed that reversible work done by the system is the maximum obtainable work. w_(rev) gt w_(ir r) The works of isothermal and adiabatic processes are different from each other. for isothermal reversible proces, W_("isothermal reversible")=2.303 nRT log_(10) (V_(2)/V_(1)) W_("adiabatic reversible")=C_(V) (T_(1)-T_(2)) A thermodynamic system goes in acyclic process as represented in the following P-V diagram: 1. For an ideal gas, consider only P-V work in going from an initial state The net work done during the complete cycle is given by the area: