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# Physics 2 - isothermal expansion?

problem #49 on this page

http://faculty.jsd.claremont.edu/sjensen...

the question is:

a 1.00 mol sample of an ideal monatomic gas is taken through the cycle shown in the figure. the process a -> b is a reversible isothermal expansion. Calculate the

a) net work done by the gas

b) the energy added to the gas by heat

c) the energy exhausted from the gas by heat

d) the efficiency of the cycle.

i know that an isothermal process means

change in temperature = 0, meaning temperature is constant

and change in internal energy = 0, meaning Q = -W

but i don't know how to apply these concepts to the problem! please help!

http://faculty.jsd.claremont.edu/sjensen/teaching/...

that is the correct website for the problem, not the one in the actual post

### 2 Answers

- schmisoLv 71 decade agoFavorite Answer
(a)

I tend to solve such problems in terms of work done ON the gas and heat added to the gas. It just a convention, but it helps to avoid sign errors, because the change of internal energy equals the energy transfer by work or heat TO the gas.

To get the work done by the gas just switch the sign of the work done on the gas.

For any process the work done ON the gas is given by the integral:

W = - ∫ p dV from initial to final state

In process A→B we've got an isothermal process.

That means

p∙V = n∙R∙T = constant

or

p∙V = p_A ∙ V_A

hence:

p = p_A∙V_A/V

So the work done in the process is

W_AB = - ∫ p_A∙V_A/V dV dV from V_A to V_B

= - p_A∙V_A ∙ ∫ 1//V dV dV from V_A to V_B

= - p_A∙V_A ∙ ln(V_B/V_A)

Convert to SI-units in order to get a result in Pa m³ =J

W_AB = - 5∙101325Pa ∙ 0.01m³ ∙ ln(0.05m³ / 0.01m³) = - 8154J

In process B→C we've got a constant pressure process.

So the work integral simplifies to

W = - p ∙ ∫ dV = -p∙∆V

Hence:

W_BC = - p_B ∙ (V_C - V_B)

= - 101325Pa ∙ (0.01m³ - 0.05m³)

= 4054J

In process C→A there is no volume change. So no work is done

W_CA = 0J

So the net work done one the gas in the whole cycle A→B→C→A is

W = W_AB + W_BC + W_CA

= -8154J + 4054J + 0J

= -4100J

That means the gas does 4101J of work to the surrounding per cycle.

(b) and (c)

The heat transfferred to the gas can be calculated via the internal energy change:

∆U = Q + W

<=>

Q = ∆U - W

The internal energy of an ideal gas is given by:

U = n∙Cv∙T

Molar hat capacity at constant volume of an ideal monatomic gas is:

Cv = (3/2)∙R

The temperatures can be found from ideal gas law:

p∙V = n∙R∙T

=>

T = p∙V/(n∙R)

You can calculate two temperatures explicitly, but you don't need it.

Simply substutite the expression for T into the internal energy definition:

U = (3/2)∙n∙R ∙p∙V/(n∙R) = (3/2)∙p∙V

Hence:

∆U = (3/2) ∙ ∆(p∙V)

The the internal energy change of the the processes is

∆U_AB = 0 because it is an isothermal proceass

∆U_BC = (3/2) ∙ (p_C∙V_C - p_B∙V_B)

(because p_C = p_B)

= (3/2) ∙ p_B ∙ (V_C - V_B)

= (3/2) ∙ 101325Pa ∙ (0.01m³ - 0.05m³)

= -6080J

∆U_CA = (3/2) ∙ (p_A∙V_A - p_C∙V_C)

(because V_C = V_A)

= (3/2) ∙ V_C ∙ (p_A - p_C)

= (3/2) ∙ 0.01m³ ∙ (5∙101325Pa - 101325Pa)

= +6080J

So the transferred to the gas in each step is:

Q_AB = ∆U_AB - W_AB

= 0 - (-8154J) = +8154J

Q_BC = ∆U_BC - W_BC

= -6080J - 4054J = -10134J

Q_CA = ∆U_CA - W_CA

= +6080J - 0J = +6080J

Negative sign indicates that energy is exhausted in process B→C.

The net energy added to the gas by heat is

Q_in = Q_AB + Q_CA = 8154J + 6080J = +14,234J

The net energy exhausted by the gas by heat is

Q_out = -Q_BC = 10,134J

(d)

efficeny is th ratio of useful work done by the gas t heat added to the gas, i.e.

η = -W/Q_in

= 4100J / 14,234J

= 0.288

= 28.8%

Right for the most part. For 10,134J is should be negative, and for net work done on gas, it should be positive 4100J.