An ideal gas is made to undergo the cyclic process shown in the figure below. Let Δ W depict the work done, Δ U be the change in internal energy of the gas and Q be the heat added to the gas. Sign of each of these three quantities for the whole cycle will be ( 0 refers to no change)

Step-by-step solution

Given cyclic process is Area under compression process C A is more than area under expansion process A B . So, net work done is negative. i.e. Δ W 0 Also, in a cyclic process, change in internal energy is zero. i.e. Δ U = 0 Now, by using first law of thermodynamics, we have Δ Q = 0 + Δ W we see that, Δ Q 0