Gas Laws - MCAT Chemical and Physical Foundations of Biological Systems

At constant pressure, $$\frac{V_{1}$$$}{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.

$\frac{8 liters}{303 K}$=\frac{16 $liters}{T_{2}$$}

$T_{2}$ = 606 K

Convert this value back to Celsius.

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

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

Solving for P gives us 12.4atm.

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

Using Gay-Lussac's Law, which is , we can find the change in temperature when the pressure is doubled.

Because the problem states that the original conditions were at STP, we know that pressure is 1atm and temperature is 273K. Since pressure and temperature are directly proportional, doubling pressure will also double the temperature. The final temperature will be 546K.

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:

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

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

In the problem, the volume and the number of moles are constant and the temperature and pressure are the only two variables that are changing. Using the ideal gas law we can find that temperature and pressure are directly proportional. When pressure increases by a factor of two, temperature will also increase by a factor of two.

Using the ideal gas law, , we can solve for pressure.

Using the equation and solving for P you get, .

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.

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

Now we can solve for P.

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

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

Charles's Law states that . To solve for the final volume, we simply plug in our given values to this equation.

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

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

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.

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.

The amount of gas is irrelevant. If the temperature is held constant and the volume is increased by a factor of three, the resulting pressure is decreased by a factor of three according to Boyle’s Law.

Remember to convert temperature to Kelvin when using the gas equations.

We will use Charles's Law to calculate the new volume:

Use the given temperatures and initial volume to calculate the final volume.

To solve this problem we use the combined gas law:

Use the given values for the temperature and volume, as well as the initial pressure, to solve for the final pressure.

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

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

First, convert temperature to Kelvin.

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

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

A decrease in pressure means a decrease in gas particle collisions. The only option that would not cause a decrease in collisions is adding moles of a different gas. Even though different molecules are added, there will be greater pressure as particle collisions will be more frequent.

Reducing temperature slows the gas particles, thus decreasing the frequency of collisions. Similarly, increasing the volume of the container and removing particles will cause a decrease in collisions, and subsequent pressure.

At constant pressure, $$\frac{V_{1}$$$}{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.

$\frac{8 liters}{303 K}$=\frac{16 $liters}{T_{2}$$}

$T_{2}$ = 606 K

Convert this value back to Celsius.

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

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

Solving for P gives us 12.4atm.

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

Using Gay-Lussac's Law, which is , we can find the change in temperature when the pressure is doubled.

Because the problem states that the original conditions were at STP, we know that pressure is 1atm and temperature is 273K. Since pressure and temperature are directly proportional, doubling pressure will also double the temperature. The final temperature will be 546K.