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kherson [118]
3 years ago
11

A gas occupies a volume of 85.0 liters at a pressure of 2.24 atm and a temperature of 22.5 degrees celsius. How many moles of ga

s are in the container?
Chemistry
1 answer:
White raven [17]3 years ago
8 0

Answer:

n = 7.86 mol

Explanation:

This question can be solved using the ideal gas law of PV = nRT.

Temperature must be in K, so we will convert 22.5C to 295 K ( Kelvin = C + 273).

R is the ideal gas constant of 0.0821.

(2.24atm)(85.0L) = n(0.0821)(295K)

Isolate n to get:

n = (2.24atm)(85.0L)/(0.0821)(295K)

n = 7.86 mol

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Explanation:

The given data is as follows.

       n = 1 mol,     V_{f} = 2V_{i}

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(a)    \Delta S = \frac{dQ}{dT}

And, according to the first law of thermodynamics

                \Delta E_{int} = Q - W

And, in an isothermal process the change in internal energy of the gas is zero.

Hence,    0 = Q - W

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Expression for work done in an isothermal process is as follows.

                   W = nRT ln \frac{V_{f}}{V_{i}}

As W = Q, Hence expression for Q will also be given as follows.

            Q = nRT ln \frac{V_{f}}{V_{i}}

Now,  

        \Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}

        [/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]

                      = nR ln \frac{2V_{i}}{V_{i}}

                       = nR ln 2

                        = 1 \times 8.314 \times 0.693

                        = 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

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                   = nRT ln \frac{2V_{i}}{V_{i}}

                   = nRT ln 2

           T = \frac{Q}{nR ln 2}

              = \frac{1500}{1 \times 8.314 ln 2}

              = 260.4 K

Therefore, temperature of the gas is 260.4 K.

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