Question

a gas that has a volume of 28 liters, a temperature of 318 k, and an unknown pressure has its volume increased to 34 liters and its temperature decreased to 300 k. if the pressure measured after the change was 2.0 atm, what was the original pressure of the gas? round to 3 sig figs. do not include units

Explanation:

Step1: Use the combined gas law formula

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\). We need to solve for \(P_1\).

Step2: Rearrange the formula for \(P_1\)

\(P_1=\frac{P_2V_2T_1}{V_1T_2}\)

Step3: Substitute the given values

Given \(V_1 = 28\), \(T_1=318\), \(P_2 = 2.0\), \(V_2=34\), \(T_2 = 300\).
\(P_1=\frac{2.0\times34\times318}{28\times300}\)

Step4: Calculate the value

First, calculate the numerator: \(2.0\times34\times318= 2.0\times10812 = 21624\)
Then, calculate the denominator: \(28\times300=8400\)
\(P_1=\frac{21624}{8400}\approx2.57\)

Answer:

\(2.57\)