Question

3- A cylinder with 0.4kg of Nitrogen gas at 27ºC keeps under pressure by a frictionless piston. Weight of this piston will increase pressure to 0.35atm more than ambient pressure (which is 1atm) at 27ºC. Therefore, at the beginning Nitrogen is at 1.35atm and at equilibrium with surrounding. Consider these four processes which is happening in order:
A) System is reaching equilibrium after merging into a container of steam.
B) A variable force slowly pushing piston while temperature stay constant at 0ºC and Nitrogen compressed to half of its initial volume. At this point piston will stop by stoppers.
C) Taking the system out from steam container in order to reach equilibrium with surrounding.
D) Now remove stoppers until system reaches completely to equilibrium with surrounding.
Determine Q, W, ΔU, ΔH for the system. (consider Nitrogen as ideal gas)

Answer 1

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) $\triangle \mathrm{H} \& \Delta \mathrm{U}$ are zero because they are sate functions and in a cyclic process the change in state functions are zero, $\Delta \mathrm{H}=\Delta \mathrm{U}=\underline{0}$

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

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 $\mathrm{W}=\mathrm{nRT} \ln (\mathrm{V} 2 / \mathrm{V} 1)$

It is given $V 2=V 1 / 2$

$\mathrm{W}=\mathrm{W} 2+\mathrm{W} 4=\mathrm{nRT} 2 \ln 2-\mathrm{nRT} 4 \ln 2$

$\mathrm{T} 2=0 \mathrm{C}=273 \mathrm{K} \text { and } \mathrm{T} 4=27 \mathrm{C}=300 \mathrm{K}$

$n=0.4 * 100 / 28 \text { moles }=14.28$

$\mathrm{W}=\mathrm{nR} \ln 2 *(273-300)=14.28 * 8.314 * 0.693^{*}-27=-2221.91 \text { joules }$, 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.