Question

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 to overcome the attractive potential

Answer #1

Derive an expression for the isothermal reversible expansion of
a van der Waals gas. Account physically for the way in which the
coefficients a and b appear in the expression. Using Maple, plot
the expression along with that for an ideal gas. For the van der
Waals gas, use a case first where a = 0 and b = 5.11 x
10-2 mol-1 and where a = 4.2 L2
atm mol-2 and b = 0. Take Vi = 1.0 L,...

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?

According to the ideal gas law, a 10.59 mol
sample of argon gas in a 0.8229 L
container at 495.4 K should exert a pressure of
523.2 atm. By what percent does the pressure
calculated using the van der Waals' equation differ from the ideal
pressure? For Ar gas, a = 1.345
L2atm/mol2 and b =
3.219×10-2 L/mol.
??? %
Hint: % difference = 100 × (P ideal - Pvan
der Waals) / P ideal

According to the ideal gas law, a 9.939 mol
sample of argon gas in a 0.8276 L
container at 500.6 K should exert a pressure of
493.3 atm. What is the percent difference between
the pressure calculated using the van der Waals' equation and the
ideal pressure? For Ar gas, a =
1.345 L2atm/mol2 and b =
3.219×10-2 L/mol.

Two ideal gas systems undergo reversible expansion starting from
the same P and V. At the end of the expansion, the two systems have
the same volume. The pressure in the system that has undergone
adiabatic expansion is lower than that in the system that has
undergone isothermal expansion. Explain this observation without
using equations.

A Carnot engine uses the expansion and compression of n moles of
argon gas, for which CV=(3/2)R. This engine operates
between temperatures TC and TH. During the
isothermal expansion a→b, the volume of the gas increases from
Va to Vb=2Va.
a)Calculate the work done during the isothermal expansion a→b.
Express your answer in terms of the variables n, TH, and
gas constant R.
b)Calculate the work Wbc done during the adiabatic
expansion b→c. Express your answer in terms of the...

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....

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 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...

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...

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