Use P1V1 = P2V2

P1 = 10atm; V1 = 50L

P2 = X; V2 = 25L

(10atm)(50L) = (x)(25L)

500 = 25x

x = 20

Since the gas is ideal, we can use a variation of the ideal gas law in order to find the unknown final pressure.

\(\displaystyle PV=nRT\)

Since we know that the number of moles is constant between both vessels (and R is a constant as well), we can simply compare the three factors being manipulated between the two vessels: pressure, volume, and temperature. Using a combination of Boyle's law and Charles's law, we can compare the two vessels to one another using the following equation.

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

Use the given values to solve for the final pressure.

\(\displaystyle \small \frac{(2.5atm)(2L)}{320K} = \frac{P_{2}(1L)}{250K}\)

\(\displaystyle \small P_{2} = 3.9atm\)

Ideal gas law (modified)

P₁V₁ = P₂V₂

P1 = 2 atm; V1 = 60L; P2 = ?; V2 = 30L

(2)(60) = (X)(30)

P2 = 4 atm

Since the total pressure is dependent on the number of moles in the container, we can use the ratio of moles before and after the reaction to determine the final pressure in the container. There are initially ten moles of gas in the container, eight moles of hydrogen and two moles of nitrogen.

The next step is to determine how many moles of ammonia are created in the reaction, and if there is any excess reactant left over after the reaction. Since there is a 1:3 ratio for hydrogen gas to nitrogen gas, only six of the moles of hydrogen gas will be used in order to react will all two moles of the nitrogen gas.

\(\displaystyle 2mol\ N_2*\frac{3mol\ H_2}{1mol\ N_2}=6mol\ H_2\)

This leaves two moles of excess hydrogen gas. Using stoichiometry and the molar ratios, we determine that four moles of ammonia are created in the reaction that comsumes two moles of nitrogen.

\(\displaystyle 2mol\ N_2*\frac{2mol\ NH_3}{1mol\ N_2}=4mol\ NH_3\)

Four moles of ammonia, plus the two remaining moles of hydrogen gas, results in six moles of total gas after the reaction has run to completion.

Six moles is 60% of the inital moles in the container, so the final pressure will be 60% of the initial pressure. We can solve using the ideal gas law.

\(\displaystyle \frac{P_1V_1}{n_1T_1}=\frac{P_1V_1}{n_1T_1}\rightarrow \frac{P_1}{n_1}=\frac{P_2}{n_2}\)

\(\displaystyle \frac{5atm}{10mol}=\frac{P_2}{6mol}\)

\(\displaystyle P_2=3atm\)

use PV = nRT

n = PV/ RT

P = 1 atm; V = 10 L; R = 0.0821 Latm/molK; T = 298 K (MUST switch temperature to K)

n = moles of gas

n = (1atm)(10L)/(0.0821Latm/molK)(298K)

n= 0.41 mol

Use PV = nRT

n = PV/RT

= (2L)(4atm) / (0.0821 Latm/molK)(310 K) <-- must change T into K

= 8/25.451

= 0.314 mol

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.

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