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Answer:
The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.
Explanation:
Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).
The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:
W system= -p*∆V
Where:
- W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
- p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
- ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)
In this case:
- p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
- ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)
Replacing:
W system= -1.013*10⁶ Pa* (-0.018 m³)
Solving:
W system= 18234 J
<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>
The answer is: D the box will NOT move or change. Hope this helps!
Answer:
Original temperature (T1) = - 37.16°C
Explanation:
Given:
Gas pressure (P1) = 2.75 bar
Temperature (T2) = - 20°C
Gas pressure (P2) = 1.48 bar
Find:
Original temperature (T1)
Computation:
Using Gay-Lussac's Law
⇒ P1 / T1 = P2 / T2
⇒ 2.75 / T1 = 1.48 / (-20)
⇒ T1 = (2.75)(-20) / 1.48
⇒ T1 = -55 / 1.48
⇒ T1 = - 37.16°C
Original temperature (T1) = - 37.16°C
