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Morgarella [4.7K]
3 years ago
11

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.35 to 1.55 L against a constant pressure

of 0.829 atm?
Chemistry
1 answer:
Mrac [35]3 years ago
5 0

Answer:- work = 319 J

Solution:- The volume of the gas decreases from 5.35 L to 1.55 L.

\Delta V=1.55L-5.35L  = -3.80 L

external pressure is given as 0.829 atm.

w=-P_e_x_t\Delta V

where w is representing work.

Let's plug in the values and calculate pressure-volume work:

w=-0.829atm(-3.80L)

w = 3.15 atm.L

We need to convert atm.L to J.

1 atm.L = 101.325 J

So, 3.15atm.L(\frac{101.325J}{1atm.L})

= 319 J

So, 319 J of work is involved in a chemical reaction. Positive work means the work is done on the system.

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What change in volume results if 50 mL of gas is cooled from 30 C to 4 C
11111nata11111 [884]

The relation between the volume of the gas and the temperature is established by Charles's law. With a decrease in the temperature, the volume decreases by 45.7 mL. Thus, option c is correct.

<h3>What is Charle's law?</h3>

Charle's law states the direct relation present between the temperature and the volume of the gas. The law is given as:

V₁ ÷ T₁ = V₂ ÷ T₂

Given,

V₁ = 50 mL

T₁ = 303.15 K

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Substituting the value the final volume is calculated as:

50 ÷ 303.15 = V₂ ÷ 277.15

V₂ = (50 × 277.15) ÷ 303.15

= 45.71 mL

Therefore, option c. 45.7 mL is the final volume.

Learn more about Charles law here:

brainly.com/question/16927784

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2 years ago
A sample of n2 effuses in 255 s. how long will the same size sample of cl2 take to effuse?
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255/R₂ = √(28/70.8)
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R₂ = 405.5 s

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