When the volume of a gas is
changed from 3.6 L to 15.5 L, the
temperature will change from
°C to 87°C.
Assume that the number of moles and the
pressure remain constant.
Be sure to notice that temperatures are
given in °C!

When The Volume Of A Gas Ischanged From 3.6 L To 15.5 L, Thetemperature Will Change FromC To 87C.Assume

Answers

Answer 1

Answer:

-189.55°C

Explanation:

Given data:

Initial volume of gas = 3.6 L

Final volume of gas = 15.5 L

Initial temperature = ?

Final temperature = 87°C (87+273 = 360 K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₁  = V₁T₂  /V₂

T₁   = 3.6 L × 360 K / 15.5 L

T₁   = 1296 L.K / 15.5 L

T₁  = 83.6 K

Kelvin to °C:

83.6 K - 273.15 = -189.55°C


Related Questions

Will give brainliest

Answers

Answer:

c. both gases have the same number of molecules

brainliest?

How many grams are in 2.3 x 1024 formula units of KNO3?

Answers

Answer:

390 g KNO₃

General Formulas and Concepts:Chemistry

Atomic Structure

Reading a Periodic TableUsing Dimensional AnalysisAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right Explanation:

Step 1: Define

2.3 × 10²⁴ formula units KNO₃

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g.mol

Molar Mass of KNO₃ - 39.10 + 14.01 + 3(16.00) = 101.11 g/mol

Step 3: Convert

[tex]2.3 \cdot 10^{24} \ formula \ units \ KNO_3(\frac{1 \ mol \ KNO_3}{6.022 \cdot 10^{23} \ formula \ units \ KNO_3} )(\frac{101.11 \ g \ KNO_3}{1 \ mol \ KNO_3} )[/tex] = 386.172 g KNO₃

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

386.172 g KNO₃ ≈ 390 g KNO₃

How many moles of copper are in 6,000,000 atoms of copper?

Answers

Answer: There will be 9.9632 × 10⁻¹⁸ moles of Copper in 6,000,000 atoms of Copper.

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