A horizontal piston-cylinder device encloses an ideal gas. A spring holds the pi

Question

A horizontal piston-cylinder device encloses an ideal gas. A spring holds the piston so that the initial pressure is 10 bar (absolute) and 0.25 m3. You slowly pull back the spring so that the gas expands reversibly (i.e., without friction) until the volume of the gas doubles. The product of the pressure and volume (PV) of the gas remains constant (specifically, mRT) during the process.  

a) Calculate the work (recall how compressible fluids do work) done by a system consisting of the piston, cylinder and enclosed gas in the reversible process described above.   

b) Calculate the total heat transfer for the process described in part a). (Hint: If mRT remains constant and the device doesn’t leak, what does that mean about T?)

c) While repeating the experiment, you sneeze and the spring pops out. The pressure instantly drops to 5 bar instead of being slowly reduced. The gas expands quickly, possibly with some friction. By the time you find the spring, the piston and cylinder end up exactly the same as they were for the process in part a).   Calculate the work done by the same system as for part a). (Hint: For this process, you know PV isn’t constant. The pressure drops instantly and it takes some finite time for the change in V to catch up.)

d) Calculate the heat transfer for the process described in part c)  

e) To really make part a) reversible, what is the maximum rate at which you could pull back the spring? In other words, how slowly would you have to go such that PV remains constant?

f) The piston-cylinder device is filled with saturated liquid water at 10 bar and 0.25 m3. The water is very slowly heated without friction so that volume eventually expands to 0.5 m3 while the pressure remains constant. What are work and heat for this process? Compare the work with the results from parts a and c. Is this process reversible?

Explanation / Answer

part a:

in the first process, temperature remains constant.

hence it is an isothermal process.

work done in case of isothermal process=n*R*T*ln(final volume/initial volume)

as final volume is double the initial volume==>final volume/initial volume=2

now, as n*R*T is constant, its value can be found by noting that PV=n*R*T for ideal gas and initially P*V=10 bar*0.25 m^3

=10^6 N/m^2*0.25 m^3

=25*10^4 J

hence work done=25*10^4*ln(2)=1.733*10^5 J

part b:

as temperature does not change, there is no change in internal energy of the system

hence heat transferred=work done=1.733*10^5 J

part c:

as the system changes state suddenly, it is an adiabatic process where there is no interaction with the outside and hence heat transferred to/frm the system=0

for adiabatic process, P*V^(gamma)=constant

assuming the ideal gas to be monoatomic,

gamma=5/3

hence the constant value=K=9.9213*10^4

work done=K*(final volume^(1-gamma)- initial volume^(1-gamma))/(1-gamma)

=9.9213*10^4*(0.5^(1-1.67)-0.25^(1-1.67))/(1-1.67)
=1.3926*10^5 J

part d:

heat transferred=0 J (as an adiabatic process)

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