Q1: In a totally irreversible isothermal expansion process for an ideal gas, ΔE = 0, ΔH = 0. Then ΔQ and ΔS will be

ANS:B - ΔQ = 0 , ΔS = +ve

In a totally irreversible isothermal expansion process for an ideal gas, both ΔE (change in internal energy) and ΔH (change in enthalpy) are zero. Since the process is isothermal (constant temperature), there is no change in internal energy (ΔE = 0), and since the gas is expanding irreversibly, there is no change in enthalpy (ΔH = 0). Now, let's consider the heat transfer (ΔQ) and entropy change (ΔS):

1. ΔQ = 0: Because the process is isothermal and ΔE = 0, the heat transfer must be zero. In an isothermal process, heat transfer occurs to maintain the constant temperature, but since the gas is ideal and the expansion is irreversible, there is no exchange of heat (ΔQ = 0).
2. ΔS > 0: In an irreversible expansion, entropy always increases. Entropy is a measure of disorder or randomness in a system, and irreversible processes tend to increase the disorder of the system. Therefore, the entropy change (ΔS) must be positive.