A window in a room is open. The next day, the temperature of the room has increased, but the pressure of the air has stayed the same. State and explain what has happened to the mass of air in the room

Assuming we can treat the air in the room as an ideal gas, it will obey the ideal gas equation of state:

PV=nRT(1)

where n is the number of moles of the gas.

The question tells us that the pressure is constant, and obviously the volume of the room is constant, so the only things that can vary are T and n. The question tells us that T has changed, so let's rearrange equation (1) to give n as a function of T:

n=PVR1T(2)

or since P, V and R are all constants:

n∝1T

So as the temperature in the room goes up the mass of air in the room goes down in inverse proportion.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .