Using the First Law of Thermodynamics

> first:=dU=dW+dQ; The first law of thermodynamics

> dQ:=0; dW:=-P*dV; dQ is zero since the process is adiablatic. The second equation is the definition of work.

> dU:=Cv*dT; The definition of internal energy U for an ideal gas since U is a function of only T in this case.

> P:=R*T/V(T); The ideal gas law. V must be written as a function of T so that Maple can solve a differential equation later.

> R:=Cp-Cv; Cp:=Cv*gamma; The relationship between Cp and Cv for an ideal gas and the definition of gamma.

> simplify(first); The first law again, with the above substitutions.

> eqn:=subs(dV=dT*D(V)(T),simplify(first/dT/Cv)); A simple substitution so that Maple knows that dV/dT is the same as D(V)(T).

> sol:=dsolve(eqn,V(T)); Solve the differentiable equation.

> C1:=solve(sol,_C1); Solve for the constant renamed C1

> unassign('P'): unassign('R'); V(T):=R*T/P; Need to clear the variable names P and R so that Maple will keep it in terms of the P and R. Ideal gas law.

> simplify(C1); C1 with the above substitution.

> C2:=simplify(1/(C1/R)); R and C1 are constants. So dividing C1 by R and taking the reciprocal will still be a constant.

> C3:=simplify(C2^((gamma-1)/gamma),assume=positive); gamma is assumed to be constant, so raising C2 to the (gamma-1)/gamma power will still be a constant. This yields eqn. 10.5-74 in the text.

> T:=P*M/R/rho; The ideal gas law again.

> C4:=simplify((C3*M/R)^(-gamma),assume=positive); M, R, and gamma are all constants. So multiplying C3 by M and dividing by R and then raising it to the -gamma power will yield another constant. This gives eqn. 10.5-75 in the text.

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