Question

g A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of work done on the system in Joules. (1 L.atm

Answer 1

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

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.