BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 47E

(a)

To determine

To find: The derivative dVdP by implicit function.

Expert Solution

Answer to Problem 47E

The derivative of the equation is dVdP=(nbV)V32n3ab+PV3n2aV.

Explanation of Solution

Given:

The equation is (P+n2aV2)(vnb)=nRT, where P is the pressure, V is the volume and T is the temperature of the gas and R, a and b are constants, T remains constant.

Derivative rules:

Chain rule: dydx=dydududx

Calculation:

Obtain the derivative the equation by using implicit differentiation.

The equation is (P+n2aV2)(vnb)=nRT.

Differentiate implicitly with respect to P,

ddx((P+n2aV2)(vnb))=ddx(nRT)

Apply the product rule and simplify the terms,

(Vnb)ddP(P+n2aV2)+(P+n2aV2)ddP(Vnb)=0(Vnb)[ddP(P)+ddP(n2aV2)]+(P+n2aV2)[ddP(V)ddP(nb)]=0(Vnb)[ddP(P)+n2addP(V2)]+(P+n2aV2)[ddP(V)ddP(nb)]=0(Vnb)[1+n2addP(V2)]+(P+n2aV2)[dVdP0]=0

Apply the chain rule,

(Vnb)[1+n2a(2V3dVdP)]+(P+n2aV2)dVdP=0(Vnb)[12n2aV3dVdP]+(P+n2aV2)dVdP=0(nbV)+(2n2aV3(nbV)+P+n2aV2)dVdP=0dVdP=nbV2n2aV3(nbV)+P+n2aV2

Simplify further and obtain the derivative,

dVdP=nbV2n3abV32n2aV2+n2aV2+P=nbV2n3abV3n2aV2+P=nbV2n3abV3n2aVV3+PV3V3=(nbV)V32n3ab+PV3n2aV

Therefore, the derivative of the equation is dVdP=(nbV)V32n3ab+PV3n2aV.

(b)

To determine

To find: The rate of change of volume with respect to pressure.

Expert Solution

Answer to Problem 47E

The rate of change of volume with respect to pressure is dVdP4.04L/atm.

Explanation of Solution

Given:

Volume V=10L

Pressure P=2.5atm.

Positive constants a=3.592 L2-atm/mole2 and b=0.04267 L/mole.

The value n=1 mole.

Calculation:

Form part (a), the derivative of the equation is dVdP=(nbV)V32n3ab+PV3n2aV.

Substitute n=1 mole, V=10 L, P=2.5atm, a=3.592 L2-atm/mole2 and b=0.04267 L/mole in dVdP=(nbV)V32n3ab+PV3n2aV.

dVdP=(1(0.04267)(10))(10)32(1)3(3.592)(0.04267)+(2.5)(10)3(1)2(3.592)(10)=(0.0426710)(1000)0.30654128+250035.92=(9.95733)(1000)0.30654128+250035.92=9957.332464.38654

Therefore, the rate of change of volume with respect to pressure is dVdP4.04L/atm.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!