Question

The PV diagram of an adiabatic expansion, the graph of P= CV^-γ, where C is a constant and γ = C_p/C_v.

GOAL Use the adiabatic pressure vs. volume relation to find a change in pressure and the work done on a gas.

PROBLEM A monatomic ideal gas at an initial pressure of 1.01  10⁵ Pa expands adiabatically from an initial volume of 1.50 m³, doubling its volume. (a) Find the new pressure. (b) Sketch the PV diagram and estimate the work done on the gas.

STRATEGY There isn't enough information to solve this problem with the ideal gas law. Instead, find the change in internal energy and use the given information to find the adiabatic index and the constant C for the process. For part (b), sketch the PV diagram and count boxes to estimate the area under the graph, which gives the work.

SOLUTION

(A) Find the new pressure.

First, calculate the adiabatic index.

γ =	Cp	=	(5/2)R	=	5
	Cv		(3/2)R		3

Find the constant C.

C = P₁V₁^γ = (1.01  10⁵ Pa)(1.50 m³)^5/3 = 1.99  10⁵ Pa · m⁵

The constant C is fixed for the entire process and can be used to find P₂.

C = P₂V₂^γ = P₂(3.00 m³)^5/3

1.99  10⁵ Pa · m⁵ = P₂(6.24 m⁵)

P₂ = 3.19  10⁴ Pa

(B) Estimate the work done on the gas from a PV diagram.

Count the boxes between V₁ = 1.50 m³ and V₂ = 3.00 m³ in the graph of P = (1.99  10⁵ Pa · m⁵)V^-5/3 in the PV diagram shown in the figure.

Number of boxes ≈ 17

Each box has an 'area' of 5.00  10³ J.

W ≈ -17 · 5.00  10³ J = -8.5  10⁴ J

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REMARKS The exact answer, obtained with calculus, is -8.43  10⁴ J, so our result is a very good estimate. The answer is negative because the gas is expanding, doing positive work on the environment, thereby reducing its own internal energy.

QUESTION For an adiabatic expansion between two given volumes and an initial pressure, which gas does more work, a monatomic gas or a diatomic gas? Why?

A diatomic gas does more work because its temperature, and therefore its pressure, is lowered less by the work it does.A diatomic gas does more work because of the greater mass of the molecules colliding with the walls of the container.     A diatomic gas does more work because 1 mole of diatomic gas contains more molecules.Both do the same work, because the pressure is the same at each volume, according to the ideal gas law.A monatomic gas does more work because its temperature, and therefore pressure, is lowered less by the work it does.

EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!

Repeat the preceding calculations for an ideal diatomic gas expanding adiabatically from an initial volume of 0.500 m³ to a final volume of 1.25 m³, starting at a pressure of 1.01  10⁵ Pa. (You must sketch the curve to find the work.)

P₂ =   Pa

W ≈   J
Sketch the graph of P vs. V, and count the number of boxes under the curve. What is the area of each box? What is the total area? What is the work done on the gas?

Answer 1

For diatomic gas , C_P = 7/2 , C_V = 5/2 , Hence ratio of specific heat = 7/5 = 1.4

Letus find the constant C from initial pressure P = 1.01 10⁵ Pa and V = 0.5 m³

....................(1)

Hence if final volume is 1.25 m³ , then final pressure = 3.827 10⁴ / 1.25 = 0.306 10⁵ Pa

Bu using the constant value from eqn.(1) , we can get volume at different pressure level .

These pressure and volume data is tabulated using Excel as given below

Sketch of graph is given below

Area under graph is divided into small parts so that each part will be a trapezium as shown in figure and the last area-5 is triangle.

Area-1 :- (1/2) 0.07 (1.01 +0.86 ) 10⁵ = 0.6545 10⁵J

Area-2 :- (1/2) 0.1 (0.86 +0.7) 10⁵ = 0.78 10⁵J

Area-3 :- (1/2) 0.1 (0.7 + 0.58 ) 10⁵ = 0.64 10⁵J

Area-4 :- (1/2) 0.1 ( 0.58 +0.5) 10⁵ = 0.54 10⁵J

Area-5 :- (1/2) 0.5 1.9 10⁵ = 0.475 10⁵J

Total workdone W = sum of these areas = 3.089 10⁴ J .......................(2)

Formula for adiabatic workdone W = { P_f V_f - P_iV_i} / ( 1 - )

W = [ ( 0.3061.25 - 1.01 0.5 )] 10⁵ / 0.4 = 3.062 10⁴ J ...................(3)

Workdone calculated in eqn.(2) and eqn.(3) are approximately same