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slega [8]
3 years ago
8

`I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon i

s 317.15 K and 0.959 atm, respectively. Suppose I don't want the pressure to change, but I want to the volume to go down to 0.378 L. What is the temperature that I need to reach when I cool down the balloon? To what temperature (in Celsius) must the balloon be cooled to reduce its volume to 0.378 L if the pressure doesn't change (remained constant)?
Chemistry
1 answer:
Andrew [12]3 years ago
3 0

Answer:

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 439 mL  = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

\frac{0.439}{317.15}=\frac{0.378}{T_2}

T_2=\frac{0.378\cdot \:317.15}{0.439}=273.08\ K

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15  

So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>

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Second question:

For the conservation of energy, the total amount of heat must be 0. The coin is losing heat, so it must be negative. The water is gaining heat, so it must be positive:

Qw - Qc = 0

Q = m*s*ΔT, where Q is the heat, m is the mass, s is the specif heat, and ΔT the temperature variation (final - initial). Qw is from water and Qc for the coin. The specif heat from the water is 4.184 J/gºC. At the thermal equilibrium, the final temperature must be equal for both.

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