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3 years ago

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Suppose that 33.3 J of heat is added to an ideal gas. The gas expands at a constant pressure of 1.45 104 Pa while changing its v

olume from 3.24 10-4 to 8.40 10-4 m3. The gas is not monatomic, so the relation does not apply.(a) Determine the change in the internal energy of the gas.

1 answer:

levacccp [35]3 years ago

5 0

Answer:

ΔU = 25.8 J

Explanation:

The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.

The work (W) of expansion can be calculated using the following expression:

W = -P. ΔV

where,

P is the external pressure

ΔV is the change in volume

W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J

The change in the internal energy (ΔU) is:

ΔU = Q + W

ΔU = 33.3 J + (-7.48 J) = 25.8 J

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if a gas sample has a pressure of 761 mmhg at 0.00°c by how much does the temperature have to decrease to lower the pressure to

Nady [450]

<h3>Answer:</h3>

272.43 K or -0.718°C

<h3>Explanation:</h3>

We are given;

The initial pressure,P1 as 761 mmHg

Initial temperature, T1 as 0.00°C which is equivalent to 273.15 K

Final pressure as 759 mmHg

We are required to calculate the final temperature;

According to pressure law, the pressure of a gas and absolute temperature are directly proportional at constant volume.

That is; Pα T

Therefore, at varying pressure and temperature,

$\frac{P1}{T1}=\frac{P2}{T2}$

To get final temperature;

$T2=\frac{P2T1}{P1}$

$T2=\frac{(759 mmHg)(273.15K)}{761 mmHg}$

$T2=272.43K$

Therefore, the final temperature will be 272.43 K or -0.718°C

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3 years ago

Which one would it be since covalent bond is not here :( please help meeee

timurjin [86]

Non polar covalent bond. The Cl's electronegativity is the same since it's the same atom. Therefore, they will be nonpolar

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3 years ago

slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.

pishuonlain [190]

<u>Answer:</u>

<u>For A:</u> The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

<u>For B:</u> The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

<u>For C:</u> The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

<u>Explanation:</u>

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

<u>For A:</u> Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

<u>For B:</u> Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

<u>For C:</u> Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

6 0

3 years ago

Confused as heck. please help!

Nitella [24]

Answer: that all thre water cycle and C is vaporation

Explanation:

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2 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

3 years ago

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