Gas Temperature Calculation Using PV=nRT

Dec 16, 2025 by Andrew McMorgan 41 views

Hey guys! Ever wondered how scientists figure out the temperature of a gas packed inside a cylinder? It's not magic, it's just some awesome chemistry using a super handy formula: $P V=n R T$ . This equation is your best friend when dealing with gases, and today we're going to break down how to use it to find the temperature ( $T$ ) when you've got the pressure ( $P$ ), volume ( $V$ ), and the amount of gas in moles ( $n$ ). We'll even be using a specific value for the gas constant, $R$ , which is $0.0821 rac{L ullet atm}{mol ullet K}$ . So, grab your virtual lab coats, and let's dive into a problem where we need to find the temperature of $0.750$ mol of a gas stored in a $6,850 mL$ cylinder at $2.21 atm$ . We'll walk through the steps, explain the units, and make sure you're totally comfortable with how this all works. Get ready to flex those chemistry muscles!

Understanding the Ideal Gas Law: PV=nRT

Alright, let's talk about the heart of this calculation: the Ideal Gas Law, represented by the equation $P V=n R T$ . This bad boy is fundamental in chemistry and physics for describing the behavior of ideal gases. An ideal gas is basically a theoretical gas made up of many randomly moving, non-interacting point particles. While no real gas is perfectly ideal, this law works really well for most gases under normal conditions (not too high pressure and not too low temperature). Let's break down what each letter stands for, because understanding these components is key to solving our problem. P stands for Pressure, which is the force exerted by the gas per unit area. It's often measured in atmospheres (atm), pascals (Pa), or millimeters of mercury (mmHg). In our problem, the pressure is given as $2.21 atm$ . V represents Volume, the space the gas occupies. Common units include liters (L) and milliliters (mL). Our cylinder holds $6,850 mL$ of gas. Now, n is the number of moles of the gas. A mole is just a unit that represents a specific amount of a substance – about $6.022 imes 10^{23}$ particles (Avogadro's number). We're dealing with $0.750$ mol of gas here. Then there's R, the ideal gas constant. This is a proportionality constant that links the energy scale to the temperature scale for a mole of particles. Its value depends on the units used for pressure, volume, and temperature. The value we're given is $R=0.0821 rac{L ullet atm}{mol ullet K}$ , which tells us we need to make sure our volume is in liters (L), our pressure is in atmospheres (atm), and our temperature will come out in Kelvin (K). Finally, T is the absolute temperature of the gas, and it must be in Kelvin (K) for this equation to work correctly. Our goal is to solve for this T! So, the equation $P V=n R T$ is essentially a mathematical relationship that connects these four variables for a given amount of gas. It's powerful because if you know any three, you can calculate the fourth, which is exactly what we're going to do. Make sense? Awesome!

Preparing Your Units for Calculation

Before we plug any numbers into our trusty $P V=n R T$ equation, we gotta make sure all our units are playing nicely together. This is a super common pitfall, guys, so pay close attention! The gas constant $R$ we're using is $0.0821 rac{L ullet atm}{mol ullet K}$ . Notice those units? We have Liters (L), atmospheres (atm), moles (mol), and Kelvin (K). For the equation to give us the correct answer, all our other measurements need to match these units. Let's check what we have: Pressure ( $P$ ) is given as $2.21 atm$ . Perfect! That matches the 'atm' in our R value. The number of moles ( $n$ ) is $0.750$ mol. Great, that matches the 'mol' in R. Now, the volume ( $V$ ) is given as $6,850 mL$ . Uh oh! Our R value uses Liters (L), not milliliters (mL). So, we need to convert milliliters to liters. Remember that there are $1000 mL$ in $1 L$ . To convert $6,850 mL$ to Liters, we simply divide by 1000:

$V = rac{6,850 mL}{1000 mL/L} = 6.850 L$

Awesome! Now our volume is in Liters. The last thing to check is temperature. The equation $P V=n R T$ requires temperature to be in Kelvin (K). Our problem doesn't give us a temperature to start with (we're solving for it!), but if it did (say, in Celsius), we'd need to convert it. To convert from Celsius ( $°C$ ) to Kelvin (K), you add 273.15 (or just 273 for simpler calculations). For instance, $0°C$ is $273 K$ . Since we're solving for $T$ , we expect our final answer to be in Kelvin, which is exactly what we want.

So, to recap, our values, properly unit-checked and converted, are:

Pressure ( $P$ ): $2.21 atm$
Volume ( $V$ ): $6.850 L$
Number of moles ( $n$ ): $0.750$ mol
Gas constant ( $R$ ): $0.0821 rac{L ullet atm}{mol ullet K}$

With these units all set, we're ready to rock and roll with the calculation!

Solving for Temperature (T)

Now for the main event, guys! We've got our equation $P V=n R T$ and all our values are prepped with the correct units. Our mission is to find the temperature ( $T$ ). To do that, we need to rearrange the equation to isolate $T$ .

Here's how we do it:

Start with the Ideal Gas Law: $P V=n R T$
To get $T$ by itself, we need to divide both sides of the equation by $n$ and $R$ . So, we divide both sides by $n R$ :

$rac{P V}{n R} = rac{n R T}{n R}$
The $n$ and $R$ on the right side cancel out, leaving us with:

$T = rac{P V}{n R}$

See? Super straightforward algebra! Now we just need to substitute in our values that we prepped in the last step:

$P = 2.21 atm$
$V = 6.850 L$
$n = 0.750$ mol
$R = 0.0821 rac{L ullet atm}{mol ullet K}$

Let's plug them in:

$T = rac{(2.21 atm) imes (6.850 L)}{(0.750 mol) imes (0.0821 rac{L ullet atm}{mol ullet K})}$

Now, let's calculate the numerator (the top part):

$2.21 imes 6.850 = 15.1385 ext{ (L ullet atm)}$

And the denominator (the bottom part):

$0.750 imes 0.0821 = 0.061575 ext{ ( rac{L ullet atm}{K})}$

Notice how the 'mol' units cancel out in the denominator calculation. That's a good sign!

Now, we divide the numerator by the denominator:

$T = rac{15.1385 ext{ (L ullet atm)}}{0.061575 ext{ ( rac{L ullet atm}{K})}}$

Let's do the division:

$T hickapprox 245.86 K$

When we look at the units, the $L ullet atm$ cancel out, and we are left with $1 / (1/K)$ , which simplifies to $K$ (Kelvin). This is exactly the unit we want for temperature!

Final Answer and Verification

So, after all that number crunching and unit checking, we found that the temperature of the gas is approximately $245.86 K$ . Let's look at the options provided: $2.95 K$ , $5.24 K$ , $138 K$ , and $246 K$ . Our calculated value of $245.86 K$ is extremely close to $246 K$ . This suggests that $246 K$ is the correct answer, likely after rounding.

Why is this temperature reasonable? Well, remember that Kelvin is an absolute temperature scale. $0 K$ is absolute zero, the theoretical point where all molecular motion stops. $273 K$ is the freezing point of water ( $0°C$ ), and $373 K$ is the boiling point of water ( $100°C$ ). A temperature of $246 K$ is below freezing ( $273 K$ ), meaning it's quite cold, but it's definitely not close to absolute zero. This is a plausible temperature for a gas stored under these conditions.

Let's quickly double-check our work:

Did we use the correct formula? Yes, $P V=n R T$ .
Did we convert units correctly? Yes, $6,850 mL$ became $6.850 L$ .
Did we rearrange the formula correctly? Yes, $T = rac{P V}{n R}$ .
Did we plug in the numbers correctly? Yes.
Did the units cancel out to give us Kelvin? Yes!

Everything checks out! The temperature of $0.750$ mol of a gas stored in a $6,850 mL$ cylinder at $2.21 atm$ is indeed approximately $246 K$ . It’s awesome how these basic principles can help us understand and predict the behavior of gases. Keep practicing, and you'll be a gas law guru in no time!