Question

2.) An ideal gas occupying a 2L flask at 760 torr is allowed to expand to a volume of 6,000 mL.
Calculate the final pressure in atm. (T is constant)
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Answer 1

Answer:

0.34atm

Explanation:

We are given: v1 as 2L ,P1 as 769 torr, v2 as 6000mL

that is:

v1 = 2L

P1 = 769 torr

v2 = 6000mL which is equivalent to 6L

i.e: 1L = 1000mL

? = 6000mL

(1L× 6000mL) / (1000mL) = 6L

using Boyle's law:

p1v1=p2v2

√ we are looking for p2

therefore:

p2= p1v1 / v2

> p2= (769 torr × 2L) / 6L = 256.33torr

but 1 torr = 0.00132 atm or

therefore 256.33torr equals:

(256.33 torr × 0.00132 atm) / 1torr = 0.33836atm

final pressure is approximately 0.34atm

Answer 2

Answer:

here the answer is P2=0.33atm

Explanation:

Let's solve Boyle's law for P2 and then substitute values for P1, V1 and V2

P2V2=P1V1

P2=P1V1/V2

=(760 torr)(2.0L) / 600Ml

P2=0.33atm

here the answer is P2=0.33 atm

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