Work done during quasi-static, isothermal expansion from P_i to P_f

[zenterix]

Homework Statement

(a) Show that the work done by an ideal gas during the quasi-static, isothermal expansion from an initial pressure $P_i$ to a final pressure $P_f$ is given by

$$W=nRT\ln\frac{P_f}{P_i}$$

(b) Calculate the work done when the pressure of 1 mol of an ideal gas is decreased quasi-statically from 20 to 1 atm, the temperature remaining constant at $20^{\circ} C$ ($R=8.31 J/mol\cdot deg$).

Relevant Equations

$W=-\int_{V_i}^{V_f} PdV$

I am posting this question after I thought I had easily solved the problem, but then when I checked the back of the book I saw that I was incorrect.

Here is what I did.

(a)

$$W=-\int_{V_i}^{V_f} PdV\tag{1}$$

$$V=V(P,T)-\frac{nRT}{P}\tag{2}$$

$$dV=\left ( \frac{\partial V}{\partial P}\right )_T dP+\left ( \frac{\partial V}{\partial T} \right )_P dT\tag{3}$$

$$=\frac{-nRT}{P^2}dP\tag{4}$$

Now we plug (4) into (1)

$$W=\int_{P_i}^{P_f} \frac{nRT}{P}dP=nRT\ln{\frac{P_f}{P_i}}\tag{5}$$

(b)

At this point I simply plugged the values into (5)

$$W=1\text{mol}\cdot 8.31 \mathrm{\frac{J}{mol ^\circ C}}\cdot 20^{\circ} \text{C}\cdot \ln{\frac{1\text{atm}}{20\text{atm}}}\tag{6}$$

$$=-7294\text{J}\tag{7}$$

The book answer says $-3.17\times 10^3$ J.

 

[DrClaude]

(b)

At this point I simply plugged the values into (5)

$$W=1\text{mol}\cdot 8.31 \mathrm{\frac{J}{mol ^\circ C}}\cdot 20^{\circ} \text{C}\cdot \ln{\frac{1\text{atm}}{20\text{atm}}}\tag{6}$$

$$=-7294\text{J}\tag{7}$$

The book answer says $-3.17\times 10^3$ J.

Be careful: indicate that you converted the 20 °C to kelvin. Otherwise, I get the same answer as you.

 

[Chestermiller]

$$W=-\int_{V_i}^{V_f} PdV\tag{1}$$

$$V=V(P,T)-\frac{nRT}{P}\tag{2}$$

Where did you get Eqn. 2 from?

Incidentally, you could have done the initial derivation differently by using the product rule for differentiation: $$PdV=d(PV)-VdP$$At constant temperature, d(PV)=nRdT=0, so, at constant temperature, $$PdV=-VdP=-\frac{nRT}{P}dp$$

 

[zenterix]

Be careful: indicate that you converted the 20 °C to kelvin. Otherwise, I get the same answer as you.

Indeed, I did use Kelvin.

 

[zenterix]

Where did you get Eqn. 2 from?

Incidentally, you could have done the initial derivation differently by using the product rule for differentiation: $$PdV=d(PV)-VdP$$At constant temperature, d(PV)=nRdT=0, so, at constant temperature, $$PdV=-VdP=-\frac{nRT}{P}dp$$

Equation 2 comes from the equation of state for ideal gases. I think you may be asking because of the typo I have in that equation.

It should be

$$V=V(P,T)=\frac{nRT}{P}$$

I accidentally had a minus sign in place of the second equals sign.