Question

A gas expands at a constant pressure of 3 atm from a volume of 0.02 cubic meters to 0.10 cubic meters. In the process it experiences an increase in internal energy of 3.0 × 105 joules. (Note: 1 Pa = 1N/m2 and 1 atm = 1.013 × 105 Pa ) What is the work done by the gas? How much heat is absorbed by the gas? Identify the thermodynamic process.

Answer 1

a) $2.4\cdot 10^4 J$

For a gas transformation occuring at a constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where

p is the gas pressure

V_f is the final volume of the gas

V_i is the initial volume

For the gas in the problem,

$p=3 atm = 3\cdot 1.013\cdot 10^5 Pa = 3.039\cdot 10^5 Pa$ is the pressure

$V_i = 0.02 m^2$ is the initial volume

$V_f = 0.10 m^3$ is the final volume

Substituting,

$W=(3.039\cdot 10^5 Pa)(0.10 m^3-0.02m^2)=24312 J = 2.4\cdot 10^4 J$

b) $3.24\cdot 10^5 J$

The heat absorbed by the gas can be found by using the 1st law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed

W is the work done

Here we have

$\Delta U = 3.0\cdot 10^5 J$

$W=2.4\cdot 10^4 J$

So we can solve the equation to find Q:

$Q=\Delta U + W = 3.0\cdot 10^5 J +2.4\cdot 10^4 J = 3.24\cdot 10^5 J$

And this process is an isobaric process (=at constant pressure).