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:\(\displaystyle PV=nRT\).

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

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

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

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

We can calculate the number of moles using the ideal gas law:

\(\displaystyle \textup{\textit{PV=nRT}}\) 

Units always have to be in SI units: atm, liters, kelvin, etc. For this question, we have SI units for everything except for the pressure, which is given in torr, rather than atm.

\(\displaystyle 760torr * \frac{1atm}{760torr} = 1atm\)
Now that we have the proper units, we can use our given values for pressure, temperature, and volume to find the moles of gas present.

\(\displaystyle (1atm)(16L)=n(0.0821\frac{Latm}{molK})(250K)\)

\(\displaystyle n=\frac{(1atm)(16L)}{(0.0821\frac{Latm}{molK})(250K)}\)

\(\displaystyle n=0.78mol\)

First, one must find the initial pressure by plugging the initial conditions into the ideal gas equation.

\(\displaystyle PV=nRT\)

\(\displaystyle P_{initial}(3L)=(2mol)(0.0821\frac{atm*L}{mol*K} )(298K)\)

\(\displaystyle P_{initial}=16.31 atm\)

Decreasing this by 4 atm leads to a final pressure of 12.31atm, which one can plug into the ideal gas equation to find the final volume.

\(\displaystyle P_{initial}-4atm=P_{final}=16.31atm-4atm=12.31atm\)

\(\displaystyle (12.31 atm)(V_{final})=(2mol)(0.0821\frac{atm*L}{mol*K} )(298K)\)

\(\displaystyle V_{final}=3.97 L\)

Because only the pressure and volume change, one can also use Boyle's law after finding the initial and final pressure values.

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

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.

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

The system is isothermal, which means the temperature is constant.

\(\displaystyle P_{i}V_{i}=nRT=P_{f}V_{f}\)

\(\displaystyle V_{f}=\frac{P_{i}V_{i}}{P_{f}}=\frac{10\:atm*35\:L}{15\:atm}=23.33\:L\rightarrow23\:L\)

\(\displaystyle \small PV = nRT\)

\(\displaystyle \small P\) is pressure, \(\displaystyle \small V =\) volume, \(\displaystyle \small n =\) moles, \(\displaystyle \small R =\) gas constant, \(\displaystyle \small T =\) temperature. Rearrange the equation, plug in the appropriate values and solve.

\(\displaystyle n=\frac{PV}{RT}\)

\(\displaystyle n=\frac{3\cdot 15}{0.08206\cdot 298}\)

\(\displaystyle n=1.84mol\)

\(\displaystyle 1.84mol \frac{32g}{1mol O_2}=58.9g\) 

\(\displaystyle \small PV = nRT\)

\(\displaystyle \small P\) is pressure, \(\displaystyle \small V =\) volume, \(\displaystyle \small n =\) moles, \(\displaystyle \small R =\) gas constant, \(\displaystyle \small T =\) temperature. Rearrange the equation, plug in the appropriate values and solve.

\(\displaystyle n=\frac{PV}{RT}\)

\(\displaystyle n=\frac{6\cdot 7}{0.08206\cdot 283}\)

\(\displaystyle n=1.81mol\)

Since we start with 68.8g of the unknown gas, and we have 1.81mol, we can find the molecular weight.

\(\displaystyle \frac{68.8g}{1.81mol}= 38\frac{g}{mol}\)

\(\displaystyle \small PV = nRT\)

\(\displaystyle \small P\) is pressure, \(\displaystyle \small V =\) volume, \(\displaystyle \small n =\) moles, \(\displaystyle \small R =\) gas constant, \(\displaystyle \small T =\) temperature. Rearrange the equation, plug in the appropriate values and solve.

\(\displaystyle n=\frac{PV}{RT}\)

\(\displaystyle n=\frac{3\cdot 4}{0.08206\cdot 293}\)

\(\displaystyle n=0.5mol\)

Since we know we have 1.82g of the unknown gas, and we have 0.5mol, we can find the molecular weight by dividing grams by moles.

\(\displaystyle \frac{1.82g}{0.5mol}=3.64\frac{g}{mol}\)

Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of each of the individual gases in the container. Boyle's Law states that the volume of a fixed amount of gas varies inversely with the pressure at constant temperature. Charles's Law states that the temperature and volume of a gas at constant pressure are directly proportional. Avogadro's Law can also be expressed by saying that one mole of any gas at standard temperature and pressure occupies 22.4L.

The pressure exerted is dependent on molar concentration only (not mass). As the containers are all the same volume, the container with the most moles will exert the most pressure. 60g of 44g/mole CO₂ gives 1.4mol, 50g of O₂ gives 1.4mol, 46g N₂ gives 1.6mol, and 54g F₂ gives 1.4mol, 40g H2 gives 40mol.