#### To determine

**To find:** The derivative dVdP by implicit function.

#### Answer

The derivative of the equation is dVdP=(nb−V)V32n3ab+PV3−n2aV.

#### Explanation

**Given:**

The equation is (P+n2aV2)(v−nb)=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=dydu⋅dudx

**Calculation:**

Obtain the derivative the equation by using implicit differentiation.

The equation is (P+n2aV2)(v−nb)=nRT.

Differentiate implicitly with respect to *P*,

ddx((P+n2aV2)(v−nb))=ddx(nRT)

Apply the product rule and simplify the terms,

(V−nb)ddP(P+n2aV2)+(P+n2aV2)ddP(V−nb)=0(V−nb)[ddP(P)+ddP(n2aV−2)]+(P+n2aV2)[ddP(V)−ddP(nb)]=0(V−nb)[ddP(P)+n2addP(V−2)]+(P+n2aV2)[ddP(V)−ddP(nb)]=0(V−nb)[1+n2addP(V−2)]+(P+n2aV2)[dVdP−0]=0

Apply the chain rule,

(V−nb)[1+n2a(−2V−3dVdP)]+(P+n2aV2)dVdP=0(V−nb)[1−2n2aV−3dVdP]+(P+n2aV2)dVdP=0−(nb−V)+(2n2aV−3(nb−V)+P+n2aV2)dVdP=0dVdP=nb−V2n2aV−3(nb−V)+P+n2aV2

Simplify further and obtain the derivative,

dVdP=nb−V2n3abV−3−2n2aV−2+n2aV−2+P=nb−V2n3abV−3−n2aV−2+P=nb−V2n3abV−3−n2aVV−3+PV3V−3=(nb−V)V32n3ab+PV3−n2aV

Therefore, the derivative of the equation is dVdP=(nb−V)V32n3ab+PV3−n2aV.

#### To determine

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

#### Answer

The rate of change of volume with respect to pressure is dVdP≈−4.04 L/atm.

#### Explanation

**Given:**

Volume V=10L

Pressure P=2.5 atm.

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=(nb−V)V32n3ab+PV3−n2aV.

Substitute n=1 mole, V=10 L, P=2.5 atm, a=3.592 L2-atm/mole2 and b=0.04267 L/mole in dVdP=(nb−V)V32n3ab+PV3−n2aV.

dVdP=(1(0.04267)−(10))(10)32(1)3(3.592)(0.04267)+(2.5)(10)3−(1)2(3.592)(10)=(0.04267−10)(1000)0.30654128+2500−35.92=(−9.95733)(1000)0.30654128+2500−35.92=−9957.332464.38654

Therefore, the rate of change of volume with respect to pressure is dVdP≈−4.04 L/atm.