The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process form an initial state `(P_1,V_1,T)` to the final state `(P_2,V_2 lt T)` is equal to

One mole of hydrogen, assumed to be ideal, is adiabatically expanded from its initial state (P_(1), V_(1), T_(1)) to the final state (P_(2), V_(2), T_(2)) . The decrease in the internal energy of the gas during this process will be given by

One mole of hydrogen, assumed to be ideal, is adiabatically expanded from its initial state (P_(1), V_(1), T_(1)) to the final state (P_(2), V_(2), T_(2)) . The decrease in the internal energy of the gas during this process will be given by

One mole of hydrogen, assumed to be ideal, is adiabatically expanded from its initial state (P_(1), V_(1), T_(1)) to the final state (P_(2), V_(2), T_(2)) . The decrease in the internal energy of the gas during this process will be given by

When a sample of ideal gas is changed from an initial state to a final state, various curves can be plotted for the process like P-V curve, V-T curve,P-T curve etc. For example, P-V curve for a fixed amount of an ideal gas at constant temperature is a rectangular hyperbola, V-T curve for a fixed amount of an ideal gas at constant volume is again a straight line. However, the shapes may vary if the constant parameters are also changed. Now, answer the following questions : Two moles of an ideal gas is changed from its initial state (16 atm, 6L) to final state (4 atm, 15L) in such a way that this change can be represented by a straight line in P-V curve.The maximum temperature attained by the gas during the above change is :(Take : R=1/12L atm K^(-1) "mol"^(-1) )

If n moles of an ideal gas are expanded is isothermally and reversibly from an initial state in which it has pressure p_(1) and volume V_(1) to the final state of volume V_(2) and pressure P_(2) , then

An ideal gas passes from one equllibrium state (P_(1) ,V_(1), T_(1), N) to anoter equilibrium state (2P_(1) , 3V_(1), T_(2), N) Then