   Chapter 12, Problem 30P

Chapter
Section
Textbook Problem

One mole of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has an internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J. For the paths IAF, IBF, and IF in Figure P12.30, calculate (a) the work done on the gas and (b) the net energy transferred to the gas by heat in the process. Figure P12.30

(a)

To determine
The work done along the path IAF.

Explanation

Section 1:

To determine: The work done along the path IA.

Answer: Thus, the work done is zero.

Explanation:

Given Info:

Constant volume of the gas is 0.300L .

Initial pressure is 2 atm.

Finial pressure is 1.50 atm.

Formula to calculate the work done of the gas is,

W=P(VfVi)

• P is the pressure
• Vf is the final volume
• Vi is the initial volume

Since, the change in volume is zero; the work done on the gas is zero.

W=0

Thus, the work done is zero.

Section2:

To determine: The work done along the path AF.

Answer: Thus, the work done is 76.0J .

Explanation:

Given Info:

Constant pressure of the gas is 1.50atm .

Initial volume ( Vi ) of the gas is 0.300L .

Finial volume ( Vf ) of the gas is 0.800L .

Formula to calculate the work done of the gas is,

W=P(VfVi)

• P is the pressure
• Vf is the final volume
• Vi is the initial volume

Substitute 1.50atm for P , 0.800L for Vf and 0.300L for Vi to find the work done,

W=[1.50atm(1.013×105Pa/atm)](0.800L0.300L)[103m31L]=76.0J

Thus, the work done is 76.0J .

Conclusion:

The total work done on the gas in the path IAF is the sum of the work done on the path IA and AF. Hence, the total work done on the path IAF is,

W=W1+W2

• W1 is the work done on the gas on the path IA
• W2 is the work done on the gas on the path AF

Substitute zero for W1 and 76.0J for W2 ,

W=076.0J76.0J

Therefore, the total work done on the path IAF is 76.0J

To determine: The work done along the path IF.

Answer: The work done on the gas in the path IF is 88.7J .

Explanation:

Given Info:

The width of rectangle is 0.500L .

The height of rectangle is 1.50atm .

The base of triangle is 0.500L .

The height of the triangle is 0.50atm .

The area under the graph in a PV diagram is equal in magnitude to the work done on the gas. If the finial volume is greater than the initial volume, then the work done on the gas is negative.

Formula to calculate the area is,

A=(W1×H1)+12(B1×H2)

• W1 is the width of rectangle
• H1 is the height of rectangle
• B1 is the base of triangle
• H2 is the height of triangle

Substitute 0.500L for width, 1.50atm for height for rectangle and 0.50atm for height and 0.500L for base for triangle to find A,

A={(0

(b)

To determine
The net energy transferred to the gas by heat in the path IAF

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