Question

We double both volume and absolute temperature of a given amount of an ideal gas.

The pressure becomes 4times larger
The pressure becomes 2times larger
The pressure becomes 4times smaller
The pressure doesn't change

Answer 1

In a nutshell, the pressure does not change.

Step-by-step explanation:

Let assume that amount of moles is conserved, then following relationship is constructed from the Equation of State for Ideal Gas:

`(P_(o)\cdot V_(o))/(T_(o)) = (P_(f)\cdot V_(f))/(T_(f))`

The final pressure is now cleared in the equation:

`P_(f) = \left((V_(o))/(V_(f)) \right)\cdot \left((T_(f))/(T_(o)) \right) \cdot P_(o)`

`P_(f) = \left((1)/(2) \right)\cdot (2) \cdot P_(o)`

`P_(f) = P_(o)`

In a nutshell, the pressure does not change.