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sveta [45]
3 years ago
8

A cylinder with a moveable piston contains 92g of Nitrogen. The external pressure is constant at 1.00 atm. The initial temperatu

re is 200K. When the temperature is decreased by 85 K, by putting it in a lower temperature freezer, the volume will decrease, according to the Ideal Gas Law. Calculate the work for this process. Express your answer in J. The conversion factor between liter atmospheres and joules is 101.3 J
Chemistry
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:

Work done in this process = 4053 J

Explanation:

Mass of the gas = 0.092 kg

Pressure is constant = 1 atm = 101325 pa

Initial temperature T_{1} = 200 K

Final temperature T_{2} = 200 - 85 = 115 K

Gas constant for nitrogen = 297 \frac{J}{kg k}

When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.

⇒ V ∝ T

⇒ \frac{V_{2} }{V_{1} } = \frac{T_{2} }{T_{1} } ------------ ( 1 )

From ideal gas equation P_{1} V_{1} = m R T_{1} ------ (2)

⇒ 101325 × V_{1} = 0.092 × 297 × 200

⇒ V_{1} = 0.054 m^{3}

This is the volume at initial condition.

From equation 1

⇒ \frac{V_{2} }{0.054} = \frac{200}{115}

⇒ V_{2} = 0.094 m^{3}

This is the volume at final condition.

Thus the work done is given by W = P [V_{2} - V_{1} ]

⇒ W = 101325 × [ 0.094 - 0.054]

⇒ W = 4053 J

This is the work done in that process.

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The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

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E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

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Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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