The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\dfrac{dT}{T},\dfrac{dP}{P}\right)$, which is the data we are given.

Here's the derivation.

Second Law of Thermodynamics:

$$dS = \dfrac{\delta Q}{T}$$

First Law of Thermodynamics:

$$dU = \delta Q + \delta W = \delta Q - PdV_m$$

Definition of Enthalpy:

$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$

Substitutions:

$$dH = \delta Q + PdV_m + V_mdP - PdV_m = \delta Q +V_mdP$$ $$\delta Q = dH - V_mdP$$

Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):

$$PV_m = RT$$ $$V_m = \dfrac{RT}{P}$$

At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).

$$dH = C_p dT$$

Substitute to get rid of $V_m$, since we don't have volume data.

$$dQ = C_p dT - \dfrac{RT}{P}dP$$

Substitute into the second law:

$$dS = C_p \dfrac{dT}{T} - R\dfrac{dP}{P}$$

Integrate (now we account for change in pressure:

$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \dfrac{dT}{T} - R \int_{P_1}^{P_2} \dfrac{dP}{P} $$

$$\Delta S = C_p \ln \left(\dfrac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$

Using this method, I get the correct answer after I multiple the molar entropy change by the number of moles.

Note, that if you are given volume change instead, you can use a similar derivation to reach the similar equation:

$$\Delta S = C_v \ln \left(\dfrac{T_2}{T_1}\right) + R \ln \left(\dfrac{V_{m,2}}{V_{m,1}} \right)$$

Reference:

• Nasa Glenn Research Center

The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\frac{dT}{T},\frac{dP}{P}\right)$, which is the data we are given.

Here's the derivation.

Second Law of Thermodynamics:

$$dS = \frac{\delta Q}{T}$$

First Law of Thermodynamics:

$$dU = \delta Q + \delta W = \delta Q - PdV_m$$

Definition of Enthalpy:

$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$

Substitutions:

$$dH = \delta Q + PdV_m + V_mdP - PdV_m = \delta Q +V_mdP$$ $$\delta Q = dH - V_mdP$$

Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):

$$PV_m = RT$$ $$V_m = \frac{RT}{P}$$

At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).

$$dH = C_p dT$$

Substitute to get rid of $V_m$, since we don't have volume data.

$$dQ = C_p dT - \frac{RT}{P}dP$$

Substitute into the second law:

$$dS = C_p \frac{dT}{T} - R\frac{dP}{P}$$

Integrate (now we account for change in pressure:

$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \frac{dT}{T} - R \int_{P_1}^{P_2} \frac{dP}{P} $$

$$\Delta S = C_p \ln \left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$

Using this method, I get the correct answer after I multiply the molar entropy change by the number of moles.

Note, that if you are given volume change instead, you can use a similar derivation to reach a similar equation:

$$\Delta S = C_v \ln \left(\frac{T_2}{T_1}\right) + R \ln \left(\frac{V_{m,2}}{V_{m,1}} \right)$$

Reference:

• Nasa Glenn Research Center

The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\dfrac{dT}{T},\dfrac{dP}{P}\right)$, which is the data we are given.

Here's the derivation.

Second Law of Thermodynamics:

$$dS = \dfrac{dQ}{T}$$

First Law of Thermodynamics:

$$dU = dQ + dW = dQ - PdV_m$$

Definition of Enthalpy:

$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$

Substitutions:

$$dH = dQ + PdV_m + V_mdP - PdV_m = dQ +V_mdP$$ $$dQ = dH - V_mdP$$

Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):

$$PV_m = RT$$ $$V_m = \dfrac{RT}{P}$$

At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).

$$dH = C_p dT$$

Substitute to get rid of $V_m$, since we don't have volume data.

$$dQ = C_p dT - \dfrac{RT}{P}dP$$

Substitute into the second law:

$$dS = C_p \dfrac{dT}{T} - R\dfrac{dP}{P}$$

Integrate (now we account for change in preessure:

$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \dfrac{dT}{T} - R \int_{P_1}^{P_2} \dfrac{dP}{P} $$

$$\Delta S = C_p \ln \left(\dfrac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$

Using this method, I get the correct answer after I multiple the molar entropy change by the number of moles.

Note, that if you are given volume change instead, you can use a similar derivation to reach the similar equation:

$$\Delta S = C_v \ln \left(\dfrac{T_2}{T_1}\right) + R \ln \left(\dfrac{V_{m,2}}{V_{m,1}} \right)$$

Reference:

• Nasa Glenn Research Center

The pressure changes, so we cannot use the simplified equation you use. From the perfect gas law, we can calculate the volume change, but we don't need to go there. The big problem is that in addition to the heat transfer, pressure-volume work is done. Work is easy to determine at constant pressure. When pressure changes also, it gets harder. However, let's use enthalpy to get rid of work (and thus the change in volume), leaving us with $dS=f\left(\dfrac{dT}{T},\dfrac{dP}{P}\right)$, which is the data we are given.

Here's the derivation.

Second Law of Thermodynamics:

$$dS = \dfrac{\delta Q}{T}$$

First Law of Thermodynamics:

$$dU = \delta Q + \delta W = \delta Q - PdV_m$$

Definition of Enthalpy:

$$dH = dU + d(PV_m) = dU +PdV_m + V_mdP$$

Substitutions:

$$dH = \delta Q + PdV_m + V_mdP - PdV_m = \delta Q +V_mdP$$ $$\delta Q = dH - V_mdP$$

Perfect gas equation (notice that I am using molar volume $V_m$, so we will be calculating molar entropy change):

$$PV_m = RT$$ $$V_m = \dfrac{RT}{P}$$

At constant pressure (Yes, I know pressure changes, bear with me. We'll put that back in).

$$dH = C_p dT$$

Substitute to get rid of $V_m$, since we don't have volume data.

$$dQ = C_p dT - \dfrac{RT}{P}dP$$

Substitute into the second law:

$$dS = C_p \dfrac{dT}{T} - R\dfrac{dP}{P}$$

Integrate (now we account for change in pressure:

$$\int_{S_1}^{S_2} dS = C_p \int_{T_1}^{T_2} \dfrac{dT}{T} - R \int_{P_1}^{P_2} \dfrac{dP}{P} $$

$$\Delta S = C_p \ln \left(\dfrac{T_2}{T_1}\right) - R \ln \left(\frac{P_2}{P_1} \right)$$

Using this method, I get the correct answer after I multiple the molar entropy change by the number of moles.

Note, that if you are given volume change instead, you can use a similar derivation to reach the similar equation:

$$\Delta S = C_v \ln \left(\dfrac{T_2}{T_1}\right) + R \ln \left(\dfrac{V_{m,2}}{V_{m,1}} \right)$$

Reference:

• Nasa Glenn Research Center