Question

(a) An infinitesimal amount of heat is added reversibly to a system. By combining the first and second laws, show that (dU = TdS - dW). (b) When heat is added to an ideal gas, its temperature and volume change from (T_1) and (V_1) to (T_2) and (V_2). Show that the entropy change of (n) moles of the gas is given by

a) Derive the expression (dU = TdS - dW) from the first and second laws.
b) Show the entropy change for an ideal gas undergoing a process.

Answer 1

(a) The expression dU = TdS - dW can be derived by combining the first law of thermodynamics
`(\(dU = \delta Q - \delta W\))` and the definition of entropy
`(\(dS = (\delta Q)/(T)\))`. Rearranging terms and substituting
`\(dS\)` yields
`\(dU = TdS - \delta W\)`.

(b) For an ideal gas undergoing a process, the entropy change
`(\(\Delta S\))` is given by
`\(\Delta S = nC_v\ln\left((T_2)/(T_1)\right) + nR\ln\left((V_2)/(V_1)\right)\)`, where \(n\) is the number of moles,
`\(C_v\)` is the molar heat capacity at constant volume, R is the gas constant, and T and V are the temperature and volume, respectively.

Step-by-step explanation:

(a) The first law of thermodynamics states
`\(dU = \delta Q - \delta W\)`, that dU is the change in internal energy,
`\(\delta Q\)` is the heat added to the system, and
`\(\delta W\)` is the work done by the system. The definition of entropy is
`\(dS = (\delta Q)/(T)\)`. Combining these equations and rearranging terms gives
`\(dU = TdS - \delta W\)`, which expresses the change in internal energy in terms of temperature and entropy.

(b) For an ideal gas, the molar heat capacity at constant volume (
`\(C_v\)`) is related to the gas constant (R) by
`\(C_v = R\)`. The entropy change
`(\(\Delta S\))` is then given by the formula
`\(\Delta S = nC_v\ln\left((T_2)/(T_1)\right) + nR\ln\left((V_2)/(V_1)\right)\)`, where n is the number of moles, T is the temperature, and V is the volume. This expression represents the change in entropy for an ideal gas undergoing a process with temperature and volume variations.