In an isolated system only the potential energy and the entropy energy matters. Initially the system can be cold so be at 0K. However, the potential energy of the system is not zero and it will start to relax down to the potential energy landscape performing work, lowering potential energy and increasing the entropy of the system (increasing kinetic energy or temperature). So one can ask, if we have only one particle lets say located on a slope of a parabolic valley. What will be the entropy and actually how one can define it? If we consider the particle then entropy is minimal of the initial state where particle was at the potential energy maximum and was not moving. Particle occupied, in this state, only one spot. However, when it moves in the valley with any kinetic energy, even behaving as a lossless harmonic oscillator (converting potential energy into kinetic an vise versa), the entropy of this state of the particle will be higher as in space coordinates the particle is not anymore localized but rather occupies some place in average. So the potential energy in average has been lowered as well as entropy increased, so leading to a more favorable state. So the pressure, the particle exerts in average on walls of the valley, one can define as an analogy of temperature which obviously depends on the shape of the valley and the initial potential energy of the particle.