Boyle's Law is:

\(\displaystyle P_{1} V_{1}=P_{2} V_{2}\)

The initial volume (\(\displaystyle V_{1}\)) and pressure (\(\displaystyle P_{1}\)) of the gas is given. The volume changes to a new volume (\(\displaystyle V_{2}\)). Our goal is to find the new pressure (\(\displaystyle P_{2}\)). Solving for the new pressure gives:

\(\displaystyle \frac{P_{1} V_{1}}{V_{2}}=\frac{P_{2} V_{2}}{V_{2}}\rightarrow P_{2}=\frac{P_{1} V_{1}}{V_{2}}\)

\(\displaystyle P_{2}=\frac{(1.05\, atm)(30\, mL)}{15\, mL}=2.10\, atm\)

Notice the answer has 3 significant figures.

To solve this question, we will need to use Gay-Lussac's Law:

\(\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}\)

This law shows a direct relationship between temperature and pressure: as temperature increases, pressure will increase proportionally. Mathematically, we can solve for the change in pressure by inputting the change in temperature. The second temperature is equal to twice the first temperature. For the fractions to remain equal, the second pressure must also be twice the first pressure.

\(\displaystyle \frac{2}{2}\)\(\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}\)

Since the only variable that has changed is temperature, we can use Gay-Lussac's law in order to compare pressure to temperature. Because temperature and pressure are on opposite sides of the ideal gas law, they are directly proportional to one another. In other words, as one increases, the other increases as well.

Gay-Lussac's law is written as follows:

\(\displaystyle \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\)

To use this equation, we must convert the temperature to Kelvin.

\(\displaystyle 25^oC+273=298K\)

\(\displaystyle 50^oC+273=323K\)

Now, use these temperatures and the initial pressure to solve for the final pressure.

\(\displaystyle \frac{4atm}{298K} = \frac{P_{2}}{323K}\)

\(\displaystyle P_{2} = 4.3atm\)

Gay-Lussac's law defines the direct relationship between temperature and pressure. When the parameters of a system change, Gay-Lussac's law helps us anticipate the effect the changes have on pressure and temperature.

\(\displaystyle \frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}^{}}\)

Boyle's law relates pressure and volume: \(\displaystyle P_1V_1=P_2V_2\)

Charles's law relates temperature and volume: \(\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}\)

The combined gas law takes Boyle's, Charles's, and Gay-Lussac's law and combines it into one law: \(\displaystyle \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

The ideal gas law relates temperature, pressure, volume, and moles in coordination with the ideal gas constant: \(\displaystyle PV=nRT\)

The graph shows that gas pressure varies directly with Kelvin temperature at a constant volume, as determined by Gay-Lussac. Gay-Lussac's law can be represented mathematically for a given mass of gas at a constant volume as follows:

\(\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}\)

Boyle's law shows the relationship between pressure and volume. Charles's law shows the relationship between volume and temperature. Newton's second law is not related to gases and shows the relationship between force, mass, and acceleration.

We can assume that the volume of the tires remains constant. This allows us to apply Gay-Lussac's law, which related pressure and temperature:

\(\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}\)

We know both the initial and final temperatures, but we must convert to Kelvin in order to use SI units.

\(\displaystyle 19^oC+273=292K\)

\(\displaystyle 105^oC+273=378K\)

Using these temperatures and the initial pressure, we can solve for the final pressure.

\(\displaystyle \frac{3.8atm}{292K}=\frac{P_2}{378K}\)

\(\displaystyle P_2=\frac{(3.8atm)(378K)}{292K}\)

\(\displaystyle P_2=4.9atm\)

According to Gay-Lussac's law, the attributes of pressure and temperature vary by direct proportionality. As temperature increases, pressure increases as well.

\(\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}\)

Because the mass and volume of the sample of the ideal gas are kept constant, a change in pressure causes only a direct change in the temperature. This can be derived from the following ideal gas equation:\(\displaystyle PV=nRT\).

\(\displaystyle \frac{T_{2}}{T_{1}}=\frac{P_{2}}{P_{1}}\)

\(\displaystyle 20^oC+273=293K\)

\(\displaystyle \frac{T_{final}}{293 K}=\frac{3 atm}{2 atm}\)

\(\displaystyle T_{final}=439.5 K\)

This problem requires us to substitute the moles of gas in the ideal gas law to mass over molar mass.

\(\displaystyle PV=nRT\)

\(\displaystyle n=\frac{mass}{molar\ mass}=\frac{m}M{}\)

We can then isolate for the mass of chlorine gas in the vessel.

\(\displaystyle \small PV = \frac{mRT}{M}\)

\(\displaystyle \small \frac{PVM}{RT} = m\)

Chlorine is diatomic, so its molecular weight will be twice the atomic mass.

\(\displaystyle M=2(35.5\frac{g}{mol})=71\frac{g}{mol}\)

The temperature must be expressed in Kelvin.

\(\displaystyle ^oC+273=K\rightarrow 37^oC+273=310K\)

Using these values, we can solve for the mass of chlorine gas in the vessel.

\(\displaystyle \small m = \frac{(3atm)(3L)(71\frac{g}{mol})}{(0.08206)(310K)} = 25.1g\)

Since we have information about temperature and we are interested in changes in pessure, Gay-Lussac's Law will be used. This says that

\(\displaystyle \frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

The initial temperature and pressure are given (STP) meaning a temperature of \(\displaystyle 273K\) and a pressure of \(\displaystyle 760\,mmHg\). There are other values for standard pressure but they have different units and since the problems requests an answer in units of \(\displaystyle mmHg\) we must use the value of standard pressure with those units. Solving for the final pressure gives

\(\displaystyle P_{2}=\frac{P_{1}T_{2}}{T_{1}}\)

Plugging in our variables gives

\(\displaystyle P_{2}=\frac{(760\,mmHg)(400K)}{(273K)}=1113.55\,mmHg\)

We want 3 significant figures since the temperature given has 3, so the final answer is

\(\displaystyle P_{2}=1110\,mmHg\)