A sample of xenon gas at a pressure of 704 mm Hg and a temperature of 45 °C, occupies a volume of 12.6 liters. If the gas is cooled at constant pressure to a temperature of 21 °C, the volume of the gas sample will be___ L.

Well, according to the Charles's Law, as long as the pressure remains constant, the volume of a gas is directly proportional to its temperature. So, all we have to do is some simple math! Let me put on my math clown hat real quick. 🤡

We can start by converting the temperatures to Kelvin, because in chemistry, we like to party with absolute temperatures. So, 45 °C is equal to 45 + 273 = 318 K, and 21 °C is equal to 21 + 273 = 294 K.

Now let's find out the new volume. We know that the initial volume is 12.6 L. Since the pressure is constant, we can set up a ratio like this:

(Volume 1 / Temperature 1) = (Volume 2 / Temperature 2)

Plugging in the values, we get:

(12.6 L / 318 K) = (Volume 2 / 294 K)

Now, let's solve this equation for the new volume, Volume 2:

Volume 2 = (12.6 L * 294 K) / 318 K

Calculating this clown out, we get:

Volume 2 = 11.6 L (rounded to one decimal place)

So, the volume of the gas sample when cooled to a temperature of 21 °C will be approximately 11.6 liters. Ta-da! 🎩

To find the volume of the gas sample at a temperature of 21 °C, we can use the combined gas law equation:

(P1 × V1)/T1 = (P2 × V2)/T2

Where:
P1 = initial pressure of the gas sample (704 mm Hg)
V1 = initial volume of the gas sample (12.6 L)
T1 = initial temperature of the gas sample in Kelvin (45 °C + 273.15 = 318.15 K)
P2 = final pressure of the gas sample (704 mm Hg)
V2 = final volume of the gas sample (unknown)
T2 = final temperature of the gas sample in Kelvin (21 °C + 273.15 = 294.15 K)

Let's substitute the values into the equation and solve for V2:

(704 mm Hg × 12.6 L) / 318.15 K = (704 mm Hg × V2) / 294.15 K

Simplifying the equation:

(704 × 12.6) / 318.15 = 704 × V2 / 294.15

8910.24 / 318.15 = V2

V2 ≈ 27.99 L

Therefore, the volume of the gas sample at a temperature of 21 °C is approximately 27.99 liters.

To find the final volume of the Xenon gas sample, we can use the combined gas law equation, which relates the initial and final temperatures and volumes of a gas sample while keeping the pressure constant.

The combined gas law equation is:

(P1 * V1)/T1 = (P2 * V2)/T2

Where:
P1 = Initial pressure = 704 mm Hg
V1 = Initial volume = 12.6 liters
T1 = Initial temperature = 45 °C + 273.15 (convert to Kelvin)
P2 = Final pressure = 704 mm Hg (given that the pressure is constant)
V2 = Final volume (what we need to find)
T2 = Final temperature = 21 °C + 273.15 (convert to Kelvin)

Let's plug in the values into the equation and solve for V2:

(704 mm Hg * 12.6 L) / (45 °C + 273.15 K) = (704 mm Hg * V2) / (21 °C + 273.15 K)

To solve for V2, we can cross-multiply and then divide both sides of the equation:

(704 mm Hg * V2) = (45 °C + 273.15 K) * (704 mm Hg * 12.6 L) / (21 °C + 273.15 K)

Now we can cancel out the units for mm Hg:

V2 = [(45 °C + 273.15 K) * (704 mm Hg * 12.6 L)] / [(21 °C + 273.15 K)]

We can now calculate this value:
V2 = [(318.15 K) * (704 mm Hg * 12.6 L)] / [294.15 K]

After canceling out the Kelvin units, we can solve the equation:

V2 = (318.15 K * 704 mm Hg * 12.6 L) / 294.15 K

V2 = (2,845,609.6 mm Hg·L) / 294.15

Now, let's convert the final volume from mm Hg·L to liters:

V2 = 9660.70 L

Therefore, the volume of the xenon gas sample at a temperature of 21 °C will be approximately 9660.70 liters.

since PV= kT, you want V such that

V/(273+21) = 12.6/(273+45)