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den301095 [7]
3 years ago
11

You have a balloon containing 1 L of air at STP in a vacuum chamber. What will the volume of the balloon be when you reduce the

pressure by half and increase the temperature to 373 K ?
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
5 0
P1V1/T1=P2V2/T2   P2=1/2 P1 V1=1 T1=298K

1 P1/298= (1/2) P1V2/373  cross P1

1/298=1/2V2/373
1/298=1/V2 746
v2=746/298
V2=2.5L


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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
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Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

1*1 = P*0.473

1 = 0.473\:P

0.473\:P = 1

P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

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\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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