Question

Suppose 4.00 mol of an ideal gas undergoes a reversible isothermal expansion from volume V1 to volume V2 = 8V1 at temperature T = 300 K. Find (a) the work done by the gas and (b) the entropy change of the gas. (c) If the expansion is reversible and adiabatic instead of isothermal, what is the entropy change of the gas?

Answer #1

a)

The differential work done by expansion against a pressure, P is
given by:

dw = PdV

For an ideal gas

P = nRT/V

dw_{rev} = nRT/ VdV

If the process is carried out isothermally, we can integrate this to obtain:

w_{rev} =
nRT*ln(V_{final}/V_{initial})

Here, n = 4 moles, T = 300K, and V_{final} =
8*V_{initial}, so:

w_{rev} = (4mol) x (8.314 J/(mol*K)) x (300K) x ln(8)

w_{rev} = 20746

b)

q = w.

But deltaS_{rev} = q_{rev}/T, so:

deltaS_{rev} = (20746 J)/(300K) = 69.15 J/K

c)

For part c, not that by definition, for an adiabatic process, q = 0. This is also a reversible process, so the entropy

change of the gas must also be equal to zero. All reversible, adiabatic processes are also isentropic processes.

(If the process is adiabatic but not reversible, then it is not necessarily isentropic.)

A Joule expansion refers to the expansion of a gas from volume
V1 to volume V2 against no
applied pressure, and is sometimes also called a free expansion.
There is no work done, because the P of -PdV is
zero. By insulating the system, this process can be done
adiabatically, so there is no change in heat. For an ideal gas, the
adiabatic process is also isothermal, so there is no change in
thermodynamic energy, ∆U = 0 (which is...

Exactly 1.27 moles of an ideal gas undergoes an isothermal
expansion (T = 259 K) from state A to state B and then returns to
state A by another process. The volume of the gas in state B is
three times its initial volume.
(a) For the process AB, find the work done by the gas and its
change in entropy. work = J change in entropy = J/K
(b) Find the gas's change in entropy for the process BA....

Thermodynamics Question
A Joule expansion refers to the expansion of a gas from volume
V1 to volume V2against no
applied pressure, and is sometimes also called a free expansion.
There is no work done, because the P of -PdV is
zero. By insulating the system, this process can be done
adiabatically, so there is no change in heat. For an ideal gas, the
adiabatic process is also isothermal, so there is no change in
thermodynamic energy, ∆U = 0 (which...

An ideal gas at 300 K has a volume of 15 L at a pressure of 15
atm. Calculate the:
(1)the ﬁnal volume of the system,
(2) the work done by the system,
(3) the heat entering thesystem,
(4) the change in internal energy when the gas undergoes
a.- A reversible isothermal expansion to a pressure of 10
atm
b.- A reversible adiabatic expansion to a pressure of 10
atm.

Ideal gas ethylene undergoes a reversible adiabatic compression
by which its temperature increases from T1 = 300 K to T2 = 450 K.
The molar entropy in the initial state is given as s1 = 100 J K–1
mol–1, and here, for ethylene, cp = ?T + c0 with ? = 0.1 J K–2
mol–1 and c0 = 13.1 J K–1 mol–1. Determine the change of the molar
entropy s2 – s1 and the change of the chemical potential ?2...

1 mole of a gas undergoes a mechanically reversible isothermal
expansion from an initial volume 1 liter to a final volume 10 liter
at 25oC. In the process, 2.3 kJ of heat is absorbed in the system
from the surrounding. The gas follows the following formula:
V=RTP+b where V is the molar specific volume, and Tand Pare
temperature (abosolute) and gas pressure respectively. Given R=
8.314 J/(mol.K) and b= 0.0005 m3. Evaluate the following a) Work
(include sign) b) Change...

a) Calculate delta S(system) for the reversible heating of 1 mol
of ethane from 298K to 1500 K at constant pressure. Use Cp = 5.351
+ 177.669x10-3 T – 687.01x10-7 T ^2 + 8.514x10-9 T ^3 (J/mol K).
Consider the reversible Carnot cycle discussed in class with 1 mol
of an ideal gas with Cv=3/2R as the working substance. The initial
isothermal expansion occurs at the hot reservoir temperature of
Thot=600C from an initial volume of 3.50 L to a...

7. 1.55 moles of Argon gas undergo an isothermal reversible
expansion from an initial volume of 5.00 L to 105. L at 300 K.
Calculate the work done during this process using: (a) the ideal
gas equation, and (b) the van der Waals equation of state. Van der
Waals parameters for Ar are available in the back of the book.
Compare the two results, what percentage of the work done by the
van der Waals gas arises due to having...

Consider the adiabatic, reversible expansion of a closed 1 mole
sample of monatomic ideal gas from P1 = 100 bar, V1 = 1dm3, and T1
= 1200K to V2 = 1.5 dm3.
What is the final temperature of the gas? What are the values of
ΔE, ΔS and w for the process described in the previous question? ΔE
= kJ ΔS = J/K w = kJ

A 0.505-mol sample of an ideal diatomic gas at 408 kPa and 309 K
expands quasi-statically until the pressure decreases to 150 kPa.
Find the final temperature and volume of the gas, the work done by
the gas, and the heat absorbed by the gas if the expansion is the
following.
(a) isothermal
final temperature K
volume of the gas
L
work done by the gas
J
heat absorbed
J
(b) adiabatic
final temperature K
volume of the gas L...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 4 minutes ago

asked 6 minutes ago

asked 8 minutes ago

asked 14 minutes ago

asked 19 minutes ago

asked 27 minutes ago

asked 27 minutes ago

asked 30 minutes ago

asked 35 minutes ago

asked 40 minutes ago

asked 45 minutes ago