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yan [13]
3 years ago
15

a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v

olume v2. a total of 1.7 kj of heat is added during the expansion process. what is v2? let the ideal-gas constant r = 8.314 j/(mol • k). a- 41 l b- 26 l c- 32 l d- 35 l
Physics
1 answer:
shutvik [7]3 years ago
6 0
Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = e^{0.3098}


v₂  =  19* e^{0.3098}

v₂ = 25.8999

v₂ ≈ 26 L        Option b.
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Correct question is;

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Answer:

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Explanation:

1/0.12 = (1/0.05) + (1/d')

Let's rearrange to get;

(1/d') = (1/0.12) - (1/0.05)

(1/d') = (1/(12/100)) - (1/(5/100))

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Let's multiply through by 60 to get rid of the denominators on the right side;

> (1/d') = 500 - 1200

> (1/d') = -700

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Explanation:

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Where:  

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<u />

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