A quantity of an ideal gas is compressed from state `(p_(1), V_(1))` to another state `(p_(2), V_(2))` (i) isothermally, (ii) adiabatically. In which process will the work done be greater?

An ideal gas is compressed from state (P_(1), V_(1), T) to state (P_(2), V_(2), T) isothermally. What is the change in the internal energy of the gas? Answer with reason.

An ideal gas undergoes four processes: isochoric, isobaric, isothermal and adiabatic. In which process is (ii) the work done zero?

2 mol of an ideal monatomic gas is carried from a state (P_(0), V_(0)) to a state (2P_(0), 2V_(0)) along a straight line path in a P - V diagram. The amount of heat absorbed by the gas in the process is given by

1 mol of an ideal gas is compressed to half of its initial volume (i) isothermally and (ii) adiabatically. Draw the p-V diagram in each case and hence state with reason, in which case the work done in less.

2 mole of an ideal monatomic gas is carried from a state (P_(o),V_(0)) to state (2P_(0),2V_(o)) along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

Three sample, A,B,C of a gas initially have the same volume. The volume of each sample is doubled by the following processes: (i) isothermal process on A, (ii) adiabatic process on B, and (iii) isobaric process on C. In which process will the work done be highest?