Middle School

We appreciate your visit to A cylinder with 0 4 kg of nitrogen gas at 27ºC is kept under pressure by a frictionless piston The weight of this piston increases. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A cylinder with 0.4 kg of nitrogen gas at 27ºC is kept under pressure by a frictionless piston. The weight of this piston increases the pressure to 0.35 atm more than ambient pressure (which is 1 atm) at 27ºC. Therefore, initially, nitrogen is at 1.35 atm and in equilibrium with the surroundings. Consider these four processes occurring in order:

A) The system reaches equilibrium after being immersed in a container of steam.
B) A variable force slowly pushes the piston while the temperature stays constant at 0ºC, and nitrogen is compressed to half of its initial volume. At this point, the piston is stopped by stoppers.
C) The system is taken out of the steam container to reach equilibrium with the surroundings.
D) The stoppers are removed until the system completely reaches equilibrium with the surroundings.

Determine \( Q \), \( W \), \( \Delta U \), \( \Delta H \) for the system. (Consider nitrogen as an ideal gas.)

Answer :

Solution and Explanation:

Step 1 Temperature from T =27C to T = 0C constant volume process


Step 2 At 0C volume changes from V1 to V1/2 , isothermally


Step 3 Temperature changes from 0C to 27C , constant volume process


Step 4 Volume changes from V/2 to V1 isothermally at 27C


From the description of the problem it is clear that the problem is a cyclic process , which involves 2 isothermal process at 27C and 0C and two constant volume process at volume coressponding to the volume of ideal gas at the given conditions. since it is a cyclic proess (comes back to the initial state) [tex]\triangle \mathrm{H} \& \Delta \mathrm{U}[/tex] are zero because they are sate functions and in a cyclic process the change in state functions are zero, [tex]\Delta \mathrm{H}=\Delta \mathrm{U}=\underline{0}[/tex]


The first law can be written as[tex]\text { Qnet }=\Delta \mathrm{U}+\text { Whet }[/tex] with appropriate sign convention , since [tex]\Delta \mathrm{U}=0[/tex]\


Qnet = Wnet

work is zero for the two constant volume process , contribution to work comes only from isothermal process ,and for ideal gas it is given as [tex]\mathrm{W}=\mathrm{nRT} \ln (\mathrm{V} 2 / \mathrm{V} 1)[/tex]

It is given [tex]V 2=V 1 / 2[/tex]

[tex]\mathrm{W}=\mathrm{W} 2+\mathrm{W} 4=\mathrm{nRT} 2 \ln 2-\mathrm{nRT} 4 \ln 2[/tex]


[tex]\mathrm{T} 2=0 \mathrm{C}=273 \mathrm{K} \text { and } \mathrm{T} 4=27 \mathrm{C}=300 \mathrm{K}[/tex]

[tex]n=0.4 * 100 / 28 \text { moles }=14.28[/tex]

[tex]\mathrm{W}=\mathrm{nR} \ln 2 *(273-300)=14.28 * 8.314 * 0.693^{*}-27=-2221.91 \text { joules }[/tex], so this amount of work has to be done on the system , since Q= W , Q = -2221.91 Joules of heat is to be transfered from the system.

Thanks for taking the time to read A cylinder with 0 4 kg of nitrogen gas at 27ºC is kept under pressure by a frictionless piston The weight of this piston increases. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada