Chemistry: One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas?

Because atoms are to small to study with tools so they use models simple :) hope this helps you...

One mole of an ideal gas doubles its volume in

johnnyaby69

19.01.2022

Explanation:

The given data is as follows.

n = 1 mol, $V_{f} = 2V_{i}$

Q = 1500 J, R = 8.314 J/mol k

(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

or, W = Q

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.

(b) As, Q = $nRT ln \frac{V_{f}}{V_{i}}$

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

= 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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