LAW of thermodynamicsFor a closed system/non-flow system undergoing a process, (1 - 2)Q1-2 = ΔE + W1-2 …1)E = Stored energy of a systemThis stored energy can be viewed as the sum of microscopic and macroscopic energies.⇒ Q1-2 = Δ (U + KE + PE) + W1-2⇒ Q1-2 = ΔU + ΔKE + ΔPE + W1-2 …2)For a non-flow or closed system at equilibrium, ΔKE and ΔPE are negligible,So, these 2 terms can be neglected.⇒ Q1-2 = ΔU + W1-2 …3)Also, for a perfect gas, the internal energy is a function of temperature only.i.e. dU = mCνdT …4)Calculation: Given equation isQ1-2 = W1-2 …5)But first law states that; Q1-2 = ΔU + W1-2 …6)Comparing 5) and 6)⇒ ΔU = 0 …7)But for perfect gas; dU = mCνdT, integrating both sides\(\mathop \smallint \nolimits_1^2 dU = \mathop \smallint \nolimits_1^2 m{C_\nu }dT\)⇒ U2 – U1 = mCν(T2 – T1) {Assuming constant m, Cν}ΔU = mCν(ΔT) …8)Comparing 7) and 8)⇒ ΔT = 0 {∵ m ≠ 0, Cν ≠ 0}⇒ T2 = T1 = Constant = Isothermal process Remember the PROPERTIES of perfect gases and apply these directly instead of writing first law.Study all the basic PROCESSES in DETAIL like the adiabatic process, Isobaric, isochoric etc.