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Chapter 19: Problem 52

The temperature of \(n\) moles of ideal gas is changed from \(T_{1}\) to \(T_{2}\) with pressure held constant. Show that the corresponding entropy change is \(\Delta S=n C_{p} \ln \left(T_{2} / T_{1}\right)\).

Short Answer

Expert verified
The entropy change when the temperature of \(n\) moles of an ideal gas is changed from \(T_{1}\) to \(T_{2}\) at constant pressure is \(\Delta S=n C_{p} \ln \left(T_{2} / T_{1}\right)\).

Step by step solution

01

Express heat transfer in terms of heat capacity

At constant pressure, the heat added or removed (\( Q \)) is given by \( Q = n C_p \Delta T \) where \( n \) is the number of moles, \( C_p \) is the heat capacity at constant pressure and \( \Delta T = T_2 - T_1 \) is the change in temperature.
02

Represent entropy change

Entropy change (\( \Delta S \)) for a reversible process is given by \( \Delta S = \int{dQ_{rev} / T} \). Here, we can substitute \( dQ_{rev} = n C_p dT \). So, we get \( \Delta S = n C_p \int_{T1}^{T2}{dT / T} \).
03

Solve the integral

To solve this integral, you'll need to know that the integral of \( 1/x \) from \( a \) to \( b \) is \( \ln(b/a) \). Thus, we obtain \( \Delta S = n C_p \ln \left(T_2 / T_1\right) \).

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