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bixtya [17]
3 years ago
8

calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation

Chemistry
1 answer:
balu736 [363]3 years ago
8 0

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= ?
  • V= 5.005 L
  • n= 1.255 mol
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K

Solving:

P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}

P= 5.62 atm

<u><em>The pressure is 5.62 atm.</em></u>

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Explanation:

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A given mass of air has a volume of 6.00 L at 101 kPa. At constant temperature, the pressure is decreased to 25.0 kPa. Calculate
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Answer:

24.24 L

Explanation:

Boyle’s law, also called Mariotte’s law, a relation concerning the compression and expansion of a gas at constant temperature.

This empirical relation, formulated by the physicist Robert Boyle in 1662, states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form, pv = k, a constant.

Real gases obey Boyle’s law at sufficiently low pressures, although the product pv generally decreases slightly at higher pressures, where the gas begins to depart from ideal behaviour.

As, PV = k

P₁ V₁ = P₂ V₂

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How many significant figures are in 5.40
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The pressure in car tires is often measured in pounds per square inch ( lb/in.2 ), with the recommended pressure being in the ra
Goshia [24]

when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm

<h3>Conversion scale</h3>

14.6959 lb/in² = 1 atm

<h3>Data obtained from the question</h3>
  • Pressure (in lb/in²) = 32.5 lb/in²
  • Pressure (in ATM) =?

<h3>How to convert 32.5 lb/in² to atm</h3>

14.6959 lb/in² = 1 atm

Therefore

32.5 lb/in² = 32.5 / 14.6959

32.5 lb/in² = 2.21 atm

Thus, 32.5 lb/in² is equivalent to 2.21 atm

Learn more about conversion:

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