Question

An air compressor does 2.5 × 106 joules of work in raising the pressure of a quantity of air while the temperature of the air remains constant. What is the heat added to the system? Identify the thermodynamic process.

Answer 1

Q = -2.5 × 106 joules

Process: isothermal

Step-by-step explanation:

Given: W = ‒2.5 × 106 J

Find: heat, Q

Temperature remains constant, so ∆U = 0

∆U = Q – W

Q = W

Q = -2.5 × 106 joules

Process: isothermal

Answer 2

`2.5\cdot 10^6 J`

Step-by-step explanation:

According to the 1st law of thermodynamics, we have:

`\Delta U = Q-W`

where

`\Delta U` is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system

We also know the change in internal energy of a system only depends on the change in temperature:

`\Delta U \propto \Delta T`

Since here the temperature of the air remains constant, the change in internal energy is zero:

`\Delta T=0 \rightarrow \Delta U = 0`

So the first equation becomes

`Q=W`

The work done by the system here is

`W=2.5\cdot 10^6 J`

Therefore, the heat added to the system is

`Q=W=2.5\cdot 10^6 J`