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Gnom [1K]
4 years ago
14

Calculate q, w, and ? U for the reversible, isothermal compression of 10 mol of ideal gas from 1 to 10 atm at a constant tempera

ture of 0°C.
Chemistry
1 answer:
Scorpion4ik [409]4 years ago
6 0

Answer : The value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.

Explanation : Given,

Moles of gas = 10 mole

Initial pressure of gas = 1 atm

Final pressure of the gas = 10 atm

Temperature of the gas = 0^oC=273+0=273K

According to the question, this is the case of isothermal reversible compression of gas.

As per first law of thermodynamic,

\Delta U=q+w

where,

\Delta U = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

w=-2.303nRT\log (\frac{P_1}{P_2})

where,

w = work done on the gas

n = number of moles of gas

R = gas constant = 8.314 J/mole K

T = temperature of gas

n = moles of the gas

P_1 = initial pressure of gas

P_2 = final pressure of gas

Now put all the given values in the above formula, we get the work done.

w=-2.303\times 10mole\times 8.314J/moleK\times 273K\times \log (\frac{1atm}{10atm})

w=52271.69J

And we know that, the heat is equal to the work done with opposite sign convention.

So, q=-52271.69J

Therefore, the value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.

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A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

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Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

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volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  = 0.10 M

The chemical equation for this reaction is :

\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = \mathtt{ \dfrac{0.035  \ mol}{ 1.0  \  L }} = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  \mathtt{CH_3COOH \ \ \   +  \ \ \ \ OH^-  \ \ \to \ \ CH_3COO^-  \ \ \ + \ \ \  H_2O}

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log \mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}

The pH of this solution = 4.74 + log \mathtt{\dfrac{0.135}{0.065}}

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

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