Answer:
a) [tex]W= 400 J[/tex]
b) [tex]T_{2}=600 K[/tex]
c) [tex]n=0.16 mol[/tex]
d) [tex]\Delta E_{int}=598 J[/tex]
e) [tex]Q=998 J[/tex]
Explanation:
a) Let's recall the definition of work.
[tex]W=\int^{Vf}_{V0}PdV[/tex]
Because the system is a quasi-static isobaric expansion, P is constant here, therefore:
[tex]W=P\int^{Vf}_{V0}dV=P(V_{f}-V_{0})[/tex]
[tex]W=2.0\cdot 10^{5}(4.0\cdot 10^{-3}-2.0\cdot 10^{-3})= 400 [Nm^{2}][/tex]
b) Using the ideal gas equation we have:
[tex]\frac{P_{1}V_{1}}{T_{1}}=nR[/tex] (1)
and [tex]\frac{P_{2}V_{2}}{T_{2}}=nR[/tex] (2)
We can note that n times R is a constant in (1) and (2), so we can equal those equations.
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex] (3)
Let's solve T₂ for (3), let's recall that P₁ = P₂, so they canceled out
[tex]T_{2}=T_{1}\cdot V_{2}/V_{1}[/tex]
[tex]T_{2}=300\cdot 4x10^{3}/2x10^{3}=600 K[/tex]
c) Using the equation of ideal gas we have:
[tex]\frac{P_{1}V_{1}}{RT_{1}}=n[/tex]
[tex]n=\frac{P_{1}V_{1}}{RT_{1}}=0.16 mol[/tex]
d) We can write the internal energy as a function of Cv, and as we know the Cv is 1.5R for a monoatomic gas.
[tex]\Delta E_{int}=n1.5R\Delta T[/tex]
[tex]\Delta E_{int}=0.16\cdot 1.5\cdot 8.3 \cdot (600-300)[/tex]
[tex]\Delta E_{int}=598 J[/tex]
e) Using the first law of thermodynamic, we have:
[tex]\Delta E_{int}=Q-W[/tex]
Finally,
[tex]Q=\Delta E_{int}+W=598+400=998 J[/tex]
Have a nice day!
