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lord [1]
3 years ago
12

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

the final volume for the gas, as described by Boyle’s law.
Chemistry
1 answer:
GenaCL600 [577]3 years ago
4 0

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

Given P₁ = 101 KPa

V₁ = 6 L

P₂ = 25 kPa

So, V₂ = P₁ V₁ /P₂ = 101 *6/25 = 24.24 L

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A 15.0 ml sample of gas at 10.0 degree Celsius and 760 torr changes to a pressure of 1252 torr at 35.0 degree Celsius. What is t
netineya [11]

Answer:

9.91 mL

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (torr)

P2 = final pressure (torr)

V1 = initial volume (mL)

V2 = final volume (mL)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

V1 = 15.0mL

V2 = ?

P1 = 760 torr

P2 = 1252 torr

T1 = 10°C = 10 + 273 = 283K

T2 = 35°C = 35 + 273 = 308K

Using P1V1/T1 = P2V2/T2

760 × 15/283 = 1252 × V2/308

11400/283 = 1252V2/308

Cross multiply

11400 × 308 = 283 × 1252V2

3511200 = 354316V2

V2 = 3511200 ÷ 354316

V2 = 9.91 mL

4 0
3 years ago
You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t
pychu [463]

Answer:

Increase

Explanation:

According to Gay-Lussac Law,

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Mathematical relationship:

P₁/T₁ = P₂/T₂

If the initial temperature and pressure is standard,

Pressure = 1 atm

Temperature = 273.15 K

then we increase the temperature to 400.0 K, The pressure will be,

1 atm / 273.15 K = P₂/400.0K

P₂ = 1 atm × 400.0 K / 273.15 K

P₂ = 400.0 atm. K /273.15 K

P₂ = 1.46 atm

Pressure is also increase from 1 atm to 1.46 atm.

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Answer:

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Explanation:

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