Chemistry, 22.07.2021 16:10 QuestionsAnsweredNow

pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which behaves as an ideal gas).The PV curve at right describes this process. Assume, as is typical near room temperature, vibrational modes are frozen out. a)Determine the energy transferred as work, the change in internal energy, and the energy transferred as heat in this process. b)Could youhave stillansweredthe questions in a) if the temperature was not provided on the plot

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Answer from: amortegaa805

Complete Question

Questions Diagram is attached below

* W=1142.86Joule

* Q=997.7J

* H=2140.5J

Explanation:

From the question we are told that:

Temperature T=337K

Pressure P=(60-55)Pa*10^5

Volume $V=(1.6-1.4)m^3*10^{-3}$

Generally the equation for gas Constant is mathematically given by

$\frac{P_2}{P_1}=\frac{V_1}{V_2}^n$

$\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n$

n=0.65

Therefore

Work-done

$W=\int{pdv}$

$W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}$

W=1142.86Joule

Generally the equation for internal energy is mathematically given by

$Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}$

Q=997.7J

Therefore

H=Q+W

H=997.7J-11.42.9

H=2140.5J

pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles o

Answer from: Quest

i think tornado but not 100%

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³