Answer:
a) The work done by the gas is [tex]W=-1.414\,kJ[/tex], that means, work is done on the gas, not viceversa.
b) [tex]Q=0[/tex] because it is an adiabatic process.
c)[tex]\Delta E=1.414\,kJ[/tex]
d) [tex]\Delta\overline K= 3.727\cdot 10^{-22}J[/tex] per atom
Explanation:
First, for an ideal monoatomic gas we have: [tex]E= \frac{3}{2}N\cdot R\cdot T[/tex], This result can be experimentally verified or derived from statistical mechanics.
From which we can see that:[tex]c_v=\frac{\partial e}{\partial T}=\frac{3}{2}R[/tex]
An adiabatic process is one in which no heat is transferred to or from the system, therefore the first law can be written as:
[tex]\Delta E =-W[/tex]
we can calculate [tex]\Delta E [/tex] as follows:
[tex]\Delta E= n\,c_v \, \Delta T[/tex] which follows from our first equation.
[tex]\\\Delta E= n\, \frac{3}{2}R \, \Delta T \\ \\ \Delta E= 6.3\, mol\, \frac{3}{2}\,8.314\frac{J}{K mol} \, 18K\\ \\ \Delta E=-W=1.414\,kJ[/tex]
Thus the gas doesn't do any work, on the contrary, work is done on the gas to compress it.
As a monoatomic gas can only store energy in kinetic energy form, we have that the change in internal energy is related to the change in kinetic energy.
[tex]E = n*N_a *\overline K[/tex]
So:
[tex]\frac{\Delta E}{N} =\Delta \overline K[/tex]
[tex]\Delta\overline K= \frac{1.414\, kJ}{6.3mol*N_a} =3.727\cdot 10^{-22}J[/tex]
