Charles's Law states that \(\displaystyle \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\). To solve for the final volume, we simply plug in our given values to this equation.

\(\displaystyle V_2=\frac{V_1T_2}{T_1}=\frac{(15L)(350K)}{(300K)}=17.5L\)

This law applies only for isobaric (constant pressure) changes.

At constant pressure, \(\displaystyle \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\), where the temperatures are measured in Kelvin (absolute temperature).

First, convert the given temperature (C) to Kelvin (K).

K = C + 273 = 30 + 273 = 303K

Plug the temperature and volumes into the above equation and solve for the final temperature.

\(\displaystyle \frac{8 liters}{303 K}=\frac{16 liters}{T_{2}}\)

\(\displaystyle T_{2} = 606 K\)

Convert this value back to Celsius.

C = K - 273 = 606 - 273 = 333^oC

Using the ideal gas law equation we can find that P= nRT/V. We then plug in the given values.

\(\displaystyle P=\frac{5mol*0.0821\frac{L*atm}{mol*K}*303K}{10L}\)

Solving for P gives us 12.4atm.

Note: 30^oC must be converted into Kelvin by adding 273K

Using the ideal gas law, \(\displaystyle PV=nRT\), we can solve for pressure.

\(\displaystyle P(5L)= (1mol)(0.0821 \frac{atm*L}{mol *K})(300K)\)

\(\displaystyle P= \frac{(1mol)(0.0821\frac{atm*L}{mol*K})(300K)}{5L}=\frac{24.63atm*L}{5L}=4.9atm\)

Using the equation \(\displaystyle PV=nRT\) and solving for P you get, \(\displaystyle P = \frac{nRT}{V}\).

Recall that hydrogen forms a diatomic molecule when in gas form. This should always be an assumption when working with hydrogen gas on the MCAT. When we convert 5g to moles, we must use a conversion factor of 2g/mol.

\(\displaystyle 5g\ H_2*\frac{1mol\ H_2}{2g\ H_2}=2.5mol\ H_2\)

Temperature must be converted to Kelvin. You must have this conversion memorized for the MCAT.

\(\displaystyle T=32^oC+273K=305K\)

Now we can solve for P.

\(\displaystyle P=\frac{nRT}{V}=\frac{(2.5mol)(0.0821\frac{L*atm}{mol*K})(305K)}{10L}=6.26atm\)

Since this is an isothermal change (constant temperature), this falls under Boyle's law.

\(\displaystyle P_1V_1 = P_2V_2\)

Taking this equation and solving for the new pressure (P₂) we come up with 5atm.

\(\displaystyle P_2=\frac{P_1V_1}{V_2}=\frac{(2atm)(5L)}{(2L)}=5atm\)

We can use the given values to determine how many moles of the gas have leaked out after 75 hours.

\(\displaystyle 0.01\frac{mol}{hr}*75hr=0.75mol\)

We now know that the balloon started with one mole, and that 0.75 moles have leaked. This means that 0.25 moles remain in the balloon. We now have our initial and final mole values and pressure values. We can rearrange the ideal gas law to isolate the variables we need, assuming that the temperature is constant.

\(\displaystyle PV = nRT\)

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

\(\displaystyle \frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}\)

Using our proportions, we can try to solve for the final volume. For simplicity, assume the initial pressure is 1 and the final pressure is 0.5.

\(\displaystyle \frac{(1)V_1}{1}=\frac{(0.5)V_2}{0.25}\)

\(\displaystyle V_1=2V_2\rightarrow V_2=\frac{1}{2}V_1\)

First, each value must be converted to the correct units given for the gas constant.

\(\displaystyle 560 mL *\frac{1 L}{1000 mL} = 0.56 L\)

\(\displaystyle 75 ^oC +273 = 348 K\)

Next, use the ideal gas law to solve for moles.

\(\displaystyle PV = nRT\)

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

\(\displaystyle n = \frac{(2.1 atm)(0.56 L)}{(0.0821)(348 K)} = 0.041 mol\)

This question asks for you to look at a set of conditions for gases, and determine relative pressures. The best equation to use for quick calculation and relation is the ideal gas law, given by:

\(\displaystyle PV=nRT\)

Rearranging this, and removing the constant (since it will not affect relative pressure), we can generate a proportionality of pressure to the other variables.

\(\displaystyle P=\frac{nT}{V}\)

We can use this proportionality with each option to determine their rankings by pressure.

\(\displaystyle P_I=\frac{(5mol)(273K)}{10L}\)

\(\displaystyle P_{II}=\frac{(5mol)(546K)}{(5L)}=\frac{2(5mol)(273K)}{\frac{1}{2}(10L)}=4P_I\)

\(\displaystyle P_{III}=\frac{(2.5mol)(273K)}{10L}=\frac{\frac{1}{2}(5mol)(273K)}{10L}=\frac{1}2{}P_I\)

\(\displaystyle P_{IV}=\frac{(10mol)(546K)}{5L}=\frac{2(5mol)2(273K)}{\frac{1}{2}(10L)}=8P_1\)

\(\displaystyle \frac{1}{2}P_I< P_I< 4P_I< 8P_I\)

\(\displaystyle P_{III}< P_I< P_{II}< P_{IV}\)

First, convert temperature to Kelvin.

\(\displaystyle 30^oC+273=303K\)

Next, use the ideal gas law to solve for moles.  

\(\displaystyle PV = nRT\)

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

\(\displaystyle n = \frac{(3 atm)(8.0 L)}{(0.0821)(303K)} = 0.96 mol\)

Finally, convert moles of ammonia to grams using molar mass.

\(\displaystyle 0.96mol\ NH_3*\frac{17.0g\ NH_3}{1mol\ NH_3}=16g\ NH_3\)