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

Answer:

$2.5\cdot 10^6 J$

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

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

Answer 2

Answer:

Q = -2.5 × 106 joules

Process: isothermal

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