Question

A flexible vessel contains 58.00 L of gas at a pressure of 2.55 atm. Under conditions of constant temperature and moles of gas, what is the volume of the gas when the pressure of the vessel is doubled?

Answer 1

`V_f=29.00"L"`

Explanation:

Assuming an ideal gas, we can determine what happens to the volume of the gas using the ideal gas law:

`PV=Nk_bT`

Which may be written: `(PV)/T="constant"`

as `k_b` is always constant (Boltzmann's constant), and we know that the number of moles of the gas `N` remains constant.

Then we can see that if the temperature is held constant, it must also be the case that `PV="constant"`. Therefore, the pressure and volume are said to be inversely proportional, meaning that if we increase one of the state variables, the other must decrease, and vice versa.

Given that the pressure doubles, we know that `P_f=2P_i`. It follows then that, given the above relationship, `V_f=1//2V_i`. In order for the right side of the equation to remain constant, if we double one of the quantities, we must make up for it by halving the other.

Therefore, the final volume is `1/2(58.00"L")=color(blue)(29.00"L")`.

Since we are given the initial pressure as well, we can even check our answer.

Initially, we have:

`PV=(2.55)(58.00)=147.9`

Then if we double the pressure and halve the volume:

`PV=(5.10)(29.00)=147.9`

Since the product remains constant, we can confirm that we've solved the problem.