A 1.40 mol sample of an ideal monatomic gas, originally at a pressure of 1.80 at

Question

A 1.40 mol sample of an ideal monatomic gas, originally at a pressure of 1.80 atm , undergoes a three-step process:

(1) it is expanded adiabatically from T1 = 581 K to T2 = 396 K;

(2) it is compressed at constant pressure until its temperature reaches T3;

(3) it then returns to its original pressure and temperature by a constant-volume process.

Questions:

q1) Determine T3.

q2) Calculate the change in internal energy for each process.

q3) Calculate the work done by the gas for each process.

q4) Calculate the heat added to the gas for each process.

q5) Calculate the change in internal energy, the work done by the gas, and the heat added to the gas for the complete cycle.

Explanation / Answer

You need to find pressure p first. Because process 2 (from state 2 to 3) is isobaric, this pressure is identical to p. Pressure in state 2 can be found from adiabatic condition for process 1.
An ideal gas undergoing a reversible adiabatic process satisfies the relation[1]
p^(1-)T^() = constant
is the heat capacity ratio. For a monatomic ideal gas it is:
= Cp / Cv = (5/2)R / (3/2)R = 5/3
Hence,
p^(-2/3)T^(5/3) = p^(-2/3)T^(5/3)
<=>
(p/p)^(2/3) = (T/T)^(5/3)
=>
p = p = p (T/T)^(5/2)
You don't need to to compute the numerical value of p for the following.

Temperature in state 3 can be found using ideal gas law.
pV = nRT
Since number of moles and volume ar held constant in process 3:
p/T = nR/V = constant.
<=>
p/T = p/T
=>
T = T(p/p) = 531K (396K/531K)^(5/2) = 255.03K

Part 2)

Generally change in internal energy for an ideal gas is given by:
E = nCvT
Molar heat capacity at constant volume for a monatomic ideal gas is:
Cv = (3/2)R

Hence,
E = (3/2)nR(T - T)
= (3/2) 1.40mol 8.3145J/molK (396K - 531K) = - 2357.16 J

E = (3/2)nR(T - T)
= (3/2) 1.40mol 8.3145J/molK (255.03K - 396K) = - 2460.06 J

E = (3/2)nR(T - T)
= (3/2) 1.4mol 8.3145J/molK (586K - 206K) = 4818.56J


Part 3)

Process 1 is adiabatic, that means no heat is exchanged. Since change in internal energy equals work done on the gas plus heat absorbed by it, work done on the gas equals ist change in internal energy:
W = E = - 2357.06 J

For process 2 solve the work integral
W = - p dV
Since p is constant it simplifies to
W = - p dV = - pV
For an ideal gas
W = - pV = - nRT
=>
W = - nR(T - T)
= - 1.40mol 8.3145J/molK (255.03K - 396K) = 1640.93 J

In process 3 the volume is constant. SO no work is done:
W = 0

Note: work done on surroundings in each process is the same, but opposite sign


Part 4)

W_tot = W + W + W
= - 2357.16 J + 1640.93 J + 0 J = -716.23 J

Internal energy is a state function, that means each state (p,V,T) of a pure gas has certain internal energy. So the total change in internal energy for a gas undergoing a cycle process is zero.
E_tot = 0

5)Total work done and total heat absorbed equals total change in internal energy :
W_tot + Q_tot = E_tot = 0
=>
Q_ tot = W_tot = 716.23J

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