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

Answer:

1.06 atm

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:

$\frac{P_{1}}{T_1} =\frac{P_{2}}{T_2}$

We can rearrange the law algebraically to solve for $P_{1}$.

${P_{1}} =\frac{(T_1)(P_{2} )}{T_2}$

Substitute your known variables and solve:

${P_{1}} =\frac{(348 K)(1.55 atm )}{506 K} \\ = 1.06 atm$