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.
