Using

$\displaystyle P_iV_i=P_fV_f$

Solving for $\displaystyle P_f$

$\displaystyle \frac{P_iV_i}{V_f}=P_f$

Plugging in values

$\displaystyle \frac{785*1}{.333}=P_f$

$\displaystyle P_f=2357\textup{ Torr}$

Using ideal gas law:

$\displaystyle PV=nRT$

Converting Celsius to Kelvin and plugging in values:

$\displaystyle 200*5=n*.0821*289$

$\displaystyle n=42moles$

$\displaystyle 42moles*\frac{6.02*10^{23}molecules}{1 mole}=2.54*10^{25}molecules$

We will use the ideal gas law to solve this problem:

$\displaystyle PV = nRT$

We are asked to find the change in volume, so let's rearrange for that:

$\displaystyle V = \frac{nRT}{P}$

Then applying this to the initial and final scenarios:

$\displaystyle V_1 = \frac{nRT_1}{P_1}$

$\displaystyle V_2 = \frac{nRT_2}{P_2}$

Then taking the ratio of scenario 2 to scenario 1:

$\displaystyle \frac{V_2}{V_1}= \frac{\frac{nrT_2}{P_2}}{\frac{nrT_1}{P_1}} = \frac{T_2P_1}{T_1P_2}$

Where:

$\displaystyle T_2 = 1.5T_1$

$\displaystyle P_2 = 4P_1$

Substituting these in, we get:

$\displaystyle \frac{V_2}{V_1}=\frac{1.5T_1P_1}{T_14P_1} = \frac{1.5}{4} = 0.375$

We will begin with the ideal gas law for this problem:

$\displaystyle PV = nRT$

Then rearranging for total moles:

$\displaystyle n = \frac{PV}{RT}$

$\displaystyle n = \frac{(2.3atm)(40L)}{\left ( 8.314\frac{J}{k\cdot mol}\right )(313K)}\left ( \frac{101325Pa}{atm}\right )\left ( \frac{1m^3}{1000L}\right )$

$\displaystyle n = 3.58 mol$

Then multiplying by 80% to get the number of moles of nitrogen:

$\displaystyle n_{nitrogen}=0.80n = 0.80(3.58)$

$\displaystyle n_{nitrogen} = 2.9 mol$

To determine the answer, we must do several steps with the ideal gas law:

$\displaystyle PV=nRT$

For this problem, the pressure and temperature are the only things to change. Therefore, we can rearrange the equation to account for this.

$\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$

However, we don't know what the value of P1 is just yet. To determine it, we must use the values we do know to solve for the initial pressure.

$\displaystyle \begin{align*} P&=\frac{nRT}{V} \\ &=\frac{1*0.0821*298}{15} \\ &= 1.631\ atm \end{align*}$

Now, we have P1. Using this, along with T1 and T2, we can solve for P2.

$\displaystyle \begin{align*} P_2&=\frac{T_2*P_1}{T_1} \\ &= \frac{323*1.631}{298}\\ &= 1.768\ atm \end{align*}$

In this question, we're given a situation in which a gas with certain defined parameters is undergoing an isobaric expansion. We're asked to determine how the temperature changes.

An isobaric expansion refers to a process in which a container expands while maintaining a constant pressure. In this problem, we're going to need to use the ideal gas law.

$\displaystyle PV=nRT$

Based on the process described in the question stem, we know that the number of moles of gas will remain the same. Furthermore, we know the pressure remains constant. And, of course, the ideal gas constant stays the same as well. Thus, we can lump all the constants together.

$\displaystyle \frac{V}{T}=\frac{nR}{P}$

Therefore, the only variables that will be changing during this process are the temperature and the volume. Consequently, we can set up a before and after expression.

$\displaystyle \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Since we're solving for the final temperature, $\displaystyle T_{2}$, let's go ahead and isolate that term.

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

Since we know the volume doubles, we can conclude that $\displaystyle 2V_{1}=V_{2}$. We can plug this into the equation to obtain:

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

Canceling common terms, we obtain:

$\displaystyle T_{2}=2T_{1}$

Thus, we see that the final temperature is twice as much as the initial temperature. We could have concluded this fairly easily without the math, since we know that temperature and volume are directly proportional to one another. If one doubles, the other will also double (assuming all other variables remain constant).

Using

$\displaystyle \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$\displaystyle V_1=V_2$

So

$\displaystyle \frac{P_1}{T_1}=\frac{P_2}{T_2}$

Solving for $\displaystyle T_2$

$\displaystyle T_2=T_1\frac{P_2}{P_1}$

Converting to Kelvin and plugging in values

$\displaystyle T_2=293*\frac{6}{1}$

$\displaystyle T_2=1758^\circ K=1485^\circ C$

For this question, we're given various parameters with regards to a gas, including its pressure, temperature, and density. We're asked to solve for the gas's molar mass.

To solve this problem, we'll need to make use of the ideal gas law.

$\displaystyle PV=nRT$

Using this expression, we'll need to manipulate it in order to account for the density of the gas. To do that, we'll have to define moles in an alternative way.

$\displaystyle n=\frac{mass}{molar\:mass}=\frac{m}{MM}$

Combining these two expressions....

$\displaystyle PV=(\frac{m}{MM})RT$

And rearranging further....

$\displaystyle \rho =\frac{m}{V}=\frac{P(MM)}{RT}$

Leads us to a term for density.

$\displaystyle \rho =\frac{P(MM)}{RT}$

Now, we can isolate the term for molar mass.

$\displaystyle MM=\frac{\rho RT}{P}$

Then we just need to plug in the values from the question stem to find the molar mass.

$\displaystyle MM=\frac{(40\:\frac{g}{L})(0.08206\:\frac{L\cdot atm}{mol\cdot K})(300\:K)}{2.4\:atm}$

$\displaystyle MM=410.3\:\frac{g}{mol}$