# Work done on a container, resulting in pressure change

Why does pumping gas into a rigid container (causing the pressure and temperature to increase), not mean that work has been done on the container?

Work is given by the formula : W = -P ΔV

If I go off the formula alone, ΔV=0 (as the container is rigid) and hence W=0

But if I ignore the formula and take the definition of work as "the energy it takes to move an object against a force", wouldn't pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container e.g. like pumping a bike tyre, the gas is not going to go into the container unless I pump it in?

For reference:

Question from textbook: A rigid container of constant volume is used to store compressed gas. When gas is pumped into the container, the pressure of the gas inside the container is increased and the temperature of the container also increases. Which statement is true of the work done on the container?

(a) The work is equal to the increase in the pressure inside the container.

(b) The work is equal to the increase in the temperature inside the container.

(c) The work is equal to the sum of the pressure and temperature increases.

(d) There is no work done on the container.

Answer from textbook: (d) the container does not expand, so as there is no change in volume of the container, no work is done on the container.

• Which textbook is this? Dec 27, 2019 at 18:07

Yes, the explanation

the container does not expand, so as there is no change in volume of the container, no work is done on the container.

is right as it takes into account that the container is rigid, so its volume can not be changed. Thus, there is no work done by the gas on the container.

Also, if you say

" pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container "

Yes, there is work done on the gas, but it is by the piston(or the pump).

You should see how the laws of Thermodynamics make clear about the boundary between system and its surroundings, so work has to be done by "something" on "something".
Also, In the question, it is clearly asked "Which statement is true of the work done on the container?" So, as the container's volume is unaltered, there is no work on it.

• Thank you for the response ABC Dexter. That makes sense. Jul 9, 2016 at 8:19

In the open system version of the first law, the work W is divided into two distinct parts: the work involved in pushing material into and out of the system, and everything else $W_{shaft}$. The latter is loosely referred to at the "shaft work" and the former is always conveniently lumped in as part of the enthalpy of the streams entering an leaving the system. In your problem, because the container is rigid and there is no rotating shaft present, the shaft work is zero. However, there is still the work involved in pushing material into the system, which is lumped in as part of the inlet enthalpy.