- #1
zenterix
- 708
- 84
- Homework Statement
- I am quite confused by the calculations involving an adiabatic process for an ideal gas.
- Relevant Equations
- Below I show the calculations.
Consider an ideal gas undergoing an adiabatic process.
The first law says that
$$dU=\delta Q+\delta w=\delta w=-PdV$$
since ##\delta Q=0## for an adiabatic process.
##U## is a function of any two of ##P,V##, and ##T##.
Consider ##U_1=U_1(T,V)## and ##U_2=U_2(T,P)##.
For an ideal gas we have
$$dU_1=\left (\frac{\partial U_1}{\partial T}\right )_VdT=C_VdT=-PdV=\frac{nRT}{V}dV\tag{1}$$
$$dU_2=\left (\frac{\partial U_2}{\partial T}\right )_PdT=C_PdT=-PdV=-\frac{nRT}{V}dV\tag{2}$$
Are these equations both correct?
For (1) we have
$$C_VdT=-\frac{nRT}{V}dV$$
and after integrating we reach
$$P_1V_1^{\gamma}=P_2V_2^{\gamma}=k$$
where ##k## is a constant and ##\gamma=1+\frac{R}{C_V}##.
Note the implicit assumption that ##C_V## is constant.
Can we do the same thing for (2) to reach
$$P_1V_1^{\gamma}=P_2V_2^{\gamma}=k$$
where ##k## is a constant and ##\gamma=1+\frac{R}{C_P}##?
Something is fishy here since ##C_V\neq C_P##.
The first law says that
$$dU=\delta Q+\delta w=\delta w=-PdV$$
since ##\delta Q=0## for an adiabatic process.
##U## is a function of any two of ##P,V##, and ##T##.
Consider ##U_1=U_1(T,V)## and ##U_2=U_2(T,P)##.
For an ideal gas we have
$$dU_1=\left (\frac{\partial U_1}{\partial T}\right )_VdT=C_VdT=-PdV=\frac{nRT}{V}dV\tag{1}$$
$$dU_2=\left (\frac{\partial U_2}{\partial T}\right )_PdT=C_PdT=-PdV=-\frac{nRT}{V}dV\tag{2}$$
Are these equations both correct?
For (1) we have
$$C_VdT=-\frac{nRT}{V}dV$$
and after integrating we reach
$$P_1V_1^{\gamma}=P_2V_2^{\gamma}=k$$
where ##k## is a constant and ##\gamma=1+\frac{R}{C_V}##.
Note the implicit assumption that ##C_V## is constant.
Can we do the same thing for (2) to reach
$$P_1V_1^{\gamma}=P_2V_2^{\gamma}=k$$
where ##k## is a constant and ##\gamma=1+\frac{R}{C_P}##?
Something is fishy here since ##C_V\neq C_P##.