Question

A gas container has an initial temperature of 348 K with an unknown pressure. When the temperature changes to 506 K the pressure is found to be 1.55 atm. What was the initial pressure in atm?

Answer 1

1.06 atm

Step-by-step explanation:

We are given the following variables to work with:

Initial pressure (P1): unknown

Initial temperature (T1): 348 K

Final temperature (T2): 506 K

Final pressure (P2): 1.55 atm

We are asked to find the initial pressure (P1). We want to choose a gas law equation that relates initial pressure and temperature to final pressure and temperature. Gay-Lussac's law does this:

`(P_(1))/(T_1) =(P_(2))/(T_2) \\`

We can rearrange the law algebraically to solve for
`P_(1)`.

`{P_(1)} =((T_1)(P_(2) ))/(T_2) \\`

Substitute your known variables and solve:

`{P_(1)} =((348 K)(1.55 atm ))/(506 K) \\ = 1.06 atm`