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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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kbertoldo7 kbertoldo7
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10 years ago
All pressure–volume–temperature relationships for gases can be combined into a single relationship known as the combined gas law. This expression can be used when looking at the effect of changes in two of these variables on the third as long as the amount of gas (number of moles) remains constant. To use the combined gas law properly, you must always express the temperatures in kelvins. The combined gas law can be represented as follows:

P1V1T1=P2V2T2

Part A
A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5.00 L, and the pressure is 760. mmHg. The balloon ascends to an altitude of 20 km, where the pressure is 76.0 mmHg and the temperature is −50. ∘C. What is the new volume, V2, of the balloon in liters, assuming it doesn't break or leak?
Express your answer with the appropriate units.
  V2 =    

Part B
Consider 4.90L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.60L and the temperature is increased to 35∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder.
Express your answer with the appropriate units.
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Doctor-2-BDoctor-2-B
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Part A
\(\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\)

To solve for V2, we first multiply both sides by T2:
\(\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}}={{P}_{2}}{{V}_{2}}\)

Now, we divide both sides by P2 to isolate V2:
\(\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}{{P}_{2}}}={{V}_{2}}\)

We have to convert the temperatures to Kelvin by adding 273 to each.  Therefore, T1 is 293 K and T2 is 223 K.  Substituting these values into the equation we derived:

\(\frac{760 mmHg * 5.00 L * 223 K}{293 K * 76.0 mmHg} = {{V}_{2}}\)

38.05 = V2

Since the question uses 3 sigfigs, the best answer is:
38.1 = V2

Part B
\(\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\)

To isolate for P2, we first multiply both sides by T2:
\(\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}}={{P}_{2}}{{V}_{2}}\)

Now, we divide both sides by V2 to isolate P2:
\(\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}}{{{T}_{1}}{{V}_{2}}}={{P}_{2}}\)

We have to convert the temperatures to Kelvin by adding 273 to each.  Therefore, T1 is 293 K and T2 is 308 K.  Substituting these values into the equation we derived:

\(\frac{365 mmHg * 4.90 L * 308 K}{293 K * 2.60 L} = {{P}_{2}}\)

723.1005 = P2.  Again, the question uses 3 sigfigs. Technically, we should record the answer as 723.101 (we don't assume numbers before a decimal are significant), I believe the answer they are looking for is 723 mmHg.  Therefore:

723 mmHg = P2
This verified answer contains over 380 words.
Pretty fly for a SciGuy

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2 months ago
Thank you so much
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