Answer:
Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]
Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
Explanation:
Using the Ideal Gases Law yoy have for pressure:
[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]
where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:
[tex]n_{1} = n_{2} = n [/tex]
In an isobaric process the pressure is constant so:
[tex]P_{1} = P_{2} [/tex]
[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]
[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]
[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]
Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]
[tex]V_{2} = 98,44 cm^{3}[/tex]
Replacing on the ideal gases formula the pressure at this piont is:
[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]
For Temperature the ideal gases formula is:
[tex]T = \frac{P V }{n R }[/tex]
For the second process you have that [tex]T_{2} = T_{3} [/tex] So:
[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]
[tex]P_{2} V_{2} = P_{3} V_{3} [/tex]
[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]
[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
