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Artyom0805 [142]
3 years ago
15

To what temperature in kelvin will a 2.3 L balloon have to be heated to expand to a volume of 400 L Assume the initial temperatu

re of the balloon is 25 degrees celsius.
Chemistry
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

T₂ = 51826.1 K

Explanation:

Given data:

Initial Volume = 2.3 L

Final volume = 400 L

Initial temperature = 25 °C (25+ 273 = 298 K)

Final temperature = ?

Solution:

V₁/T₁ = V₂/T₂

T₂ = V₂ T₁/V₁

T₂ =  400 L . 298 K / 2.3 L

T₂ = 119200 K. L / 2.3 L

T₂ = 51826.1 K

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Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

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Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

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Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

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