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?
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.