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lara [203]
3 years ago
10

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

loon occupy at a place where the pressure is 0.600 atm and the temperature is 253 K?
Chemistry
1 answer:
bixtya [17]3 years ago
8 0

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

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A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperat
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Answer:

8,19 L

Explanation:

The combined gas law is a equation that can be used <em>when the initial and final conditions -</em>pressure,volume,amount of moles,temperature-, <em>of a gas change</em> during a process. It can be written:

\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}

If the amount of the gas remains constant, then n1=n2 and we have:

\frac{P1V1}{T1}=\frac{P2V2}{T2}

For the problem we have:

<em>Note: When working with gases is important to use </em><em><u>absolute temperature values (°K, °K=°C+273,15):</u></em>

P1=1 atm, V1=2,5L, T1=25+273,15=298,15°K

P2=0,3 atm, V2=?, T2=20+273,15=293,15°K

V2=\frac{P1V1T2}{T1P2}=\frac{1atm*2,5L*293,15K}{298,15K*0,3atm}=8,19L

The new volume of the balloon is 8,19 L.

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