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Lady_Fox [76]
2 years ago
13

When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by 10°C. What is the work done on

the gas?
Chemistry
1 answer:
Molodets [167]2 years ago
7 0

When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by 10°C. The work done on the gas is 845J.

The amount of work done on gas depends upon the internal energy change and the heat supplied to the system.

According to First Law of Thermodynamics, the change in internal energy is equal to the work done and the heat supplied to the system.

This is given by:

ΔU = W + Q

where, ΔU is change in Internal energy

            W is the Work done

            Q is the heat supplied

Given,

Q = 400J

Number of moles, n = 10

Change in temperature, ΔT = 10°C

Cv = 3/2R ; Since, the given gas is monoatomic (R=8.3)

We know that, ΔU = n Cv ΔT

On substituting the values in above formula,

ΔU = 10 × 3/2 × 8.3 × 10

ΔU = 1245J

Using,

ΔU = W + Q

1245J = W + 400

W = 845J

Hence, the work done on the gas is 845J.

Learn more about Thermodynamics here, brainly.com/question/1368306

#SPJ4

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Protons

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Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu
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Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

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Answer:

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  • So, the right choice is:

(A) a compound that donates protons .

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