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ch4aika [34]
3 years ago
7

Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c

and the pressure is constant. as part of a machine design project, calculate the final temperature of the gas after it has done 2.40 * 103 j of work.
Chemistry
1 answer:
Alla [95]3 years ago
6 0

We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.

The final temperature is: 75.11 °C.

The work done at constant pressure, W=nR(T₂-T₁)

n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).

W=2.4 × 10³ Joule (Given)

From the expression,

(T₂-T₁)=\frac{W}{nR}

(T₂-T₁)= \frac{2.40 X 10^{3} }{6 X 8.314}

(T₂-T₁)= 48.11

T₂=300+48.11=348.11 K= 75.11 °C

Final temperature is 75.11 °C.


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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.7 × 10²⁵ molecules CaF

[Solve] moles CaF

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                          \displaystyle 1.7 \cdot 10^{25} \ molecules \ CaF(\frac{1 \ mol \ CaF}{6.022 \cdot 10^{23} \ molecules \ CaF})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 28.2298 \ moles \ CaF

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

28.2298 mol CaF ≈ 28 mol CaF

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