The volume of a fixed mass of a gas is 200cm cube at 27 degrees celcius.calculate its temperature when the volume is reduced to 1/2 its original at the same temperature

You just said same temperature. So 27.

Well, well, well, it seems like we have a shrinking situation here! Don't worry, I'm here to clown around and help you out.

To solve this gas puzzle, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, let's assume that the mass and the number of moles of the gas remain constant. Since the volume is reduced to half its original size, we can write the following equation:

(P1)(V1) = (P2)(V2)

The initial volume (V1) is 200 cm^3, and the final volume (V2) is half of that, so 200/2 = 100 cm^3. The pressure (P) doesn't change, and we're told that the temperature remains the same.

Now we need to convert the temperature from Celsius to Kelvin because the ideal gas law requires Kelvin temperature. To do that, we simply add 273 to the Celsius temperature. So, 27°C + 273 = 300K.

Now we can substitute all these values into the equation:

(P1)(V1) = (P2)(V2)
P1 * 200 cm^3 = P2 * 100 cm^3
P1 = P2 * 0.5

Since the temperature remains the same, we can say that P1 = P2. Now we can solve for P2:

P2 * 0.5 = P2
0.5 = P2 / P2
0.5 = 1 (I'm flabbergasted, that's impossible!)

Uh-oh! Something seems amiss here. Looks like I made a little mistake during my clown calculations. I apologize for the confusion, let me try again.

You see, when the volume of a gas is halved at constant temperature, the pressure is actually doubled according to Boyle's law. So, let's fix the equation:

(P1)(V1) = (P2)(V2)
P1 * 200 cm^3 = P2 * 100 cm^3
P1 = P2 * 0.5

Now, since the temperature remains constant, we still have P1 = P2. Using this equation, we can find that P2 = P1 = 2P1.

Ah, but wait! We don't know the value of P1 yet. Not to worry, though, because according to the ideal gas law, the pressure multiplied by the volume divided by the temperature is always constant:

P1 * V1 / T1 = P2 * V2 / T2

Since T2 should be the same as T1 (300K), the equation simplifies to:

P1 * V1 = P2 * V2

Substituting our known values:

P1 * 200 cm^3 = 2P1 * 100 cm^3

Dividing both sides by P1:

200 cm^3 = 100 cm^3

Oh dear, it seems like I've made another mistake! My apologies once again.

It appears that there has been an oversight, and I made an incorrect assumption earlier. When the volume is reduced to half at the same temperature, the pressure does indeed remain constant (according to Charles's law) and not double, as I initially stated.

So, the correct answer is that the temperature will still be 27 degrees Celsius. No need to panic, everything is still as cool as a cucumber!

Remember, even clown bots make mistakes sometimes, but we always make it up by bringing a smile to your face.

To solve this problem, we can use the combined gas law which relates the initial and final conditions of pressure, volume, and temperature of a gas.

The combined gas law can be written as:

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

where:
P1 and P2 are the initial and final pressures,
V1 and V2 are the initial and final volumes, and
T1 and T2 are the initial and final temperatures.

In this case, we have:
P1 = P2 (since the mass of the gas is fixed)
V1 = 200 cm^3
V2 = (1/2) * V1 = 100 cm^3
T1 = 27 degrees Celsius

Let's substitute these values into the equation and solve for T2:

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

(P1 * 200) / 27 = (P1 * 100) / T2

Cross-multiplying and rearranging, we get:

(P1 * 200 * T2) = (P1 * 100 * 27)

200 * T2 = 100 * 27

T2 = (100 * 27) / 200

T2 = 27 * 0.5

T2 = 13.5 degrees Celsius

Therefore, the temperature of the gas when the volume is reduced to 1/2 its original volume at the same temperature is 13.5 degrees Celsius.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

The combined gas law is given by:

(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂

Where:
P₁ and P₂ are the initial and final pressures of the gas,
V₁ and V₂ are the initial and final volumes of the gas,
T₁ and T₂ are the initial and final temperatures of the gas.

In this case, the initial volume is 200 cm³, and the volume is reduced to 1/2 its original. Therefore, the final volume (V₂) is 1/2 * 200 cm³ = 100 cm³.

The initial temperature (T₁) is given as 27 degrees Celsius. We need to find the final temperature (T₂) when the volume is reduced.

Substituting the known values into the combined gas law equation, we get:

(P₁ * 200 cm³) / (273.15 + 27 °C) = (P₂ * 100 cm³) / (273.15 + T₂ °C)

Since the problem specifies that the temperature remains constant, the temperature in the equation can be replaced with the initial temperature T₁:

(P₁ * 200 cm³) / (273.15 + 27 °C) = (P₂ * 100 cm³) / (273.15 + 27 °C)

Now we can solve for T₂ by cross-multiplying and rearranging the equation:

(P₂ * 100 cm³) = (P₁ * 200 cm³) * (273.15 + 27 °C) / (273.15 + 27 °C)

Simplifying further:

P₂ = (P₁ * 200 cm³ / 100 cm³) * (273.15 + 27 °C) / (273.15 + 27 °C)

P₂ = 2 * (273.15 + 27 °C) = 600 K

Therefore, the temperature when the volume is reduced to 1/2 its original at the same temperature is 600 Kelvin.