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Olegator [25]
3 years ago
11

A sample of nitrogen gas expands in volume from 1.3 to 5.7 L at constant temperature. Calculate the work done in joules if the g

as expands (a) against a vacuum: J (b) against a constant pressure of 0.60 atm: × 10 J Enter your answer in scientific notation. (c) against a constant pressure of 4.7 atm: × 10 J Enter your answer in scientific notation.
Chemistry
1 answer:
just olya [345]3 years ago
7 0

Answer:

a) 0 J

b)  -2.67x10² J

c) -2.09x10³ J

Explanation:

For an isothermic expansion (with constant temperature) the work (W) is :

W = -pΔV, where p is the pressure and ΔV the volume variation. The minus signal is used because the compression is positive and ΔV is negative (Vf < Vi).

a) In vacuum, the relative pressure is 0 atm, so the work:

W = -0x(5.7 - 1.3)

W = 0 J

b) For a constant pressure of 0.60 atm

W = -0.6atmx(5.7 - 1.3)L = -2.64 L.atm

1 L.atm = 101.3 J

W = -2.64x101.3 = -2.67x10² J

c) For a pressure of 4.7 atm

W = -4.7atmx(5.7 - 1.3) L = - 20.68 atm.L

1 atm.L = 101.3 J

W = -20.68x101.3 = -2.09x10³ J

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3 years ago
Convert 1.248×1010 g to each of the following units.
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Answer:

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d) 1.248 x 10⁴ ton

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a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):

1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg

b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):

1.248 x 10¹⁰ g * (  1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg

c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):

1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg

d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):

1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton

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+ 291.9 kJ

Solution:

The equation given is as;

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First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

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Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

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