Chemistry, 21.02.2020 05:09 ilovepickles930

3.00 mol of an ideal gas with CV, m=3R/2 undergoes an expansion from an initial state described by T=310.K and P=6.00bar against a constant external pressure of zero bar until the final pressure is equal to one-fifth of its initial value. The state of the surroundings is T=298K, P=0.250bar. a. FInd entropy of surroundings.

b. Find total entropy

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Answer from: ronniethefun

Explanation:

The given data is as follows.

n = 3 mol, $C_{v} = \frac{3R}{2}$

$P_{2} = \frac{1}{5} \times 6 bar$ > \frac{1}{5} \times P_{1}[/tex]

so, $P_{2}$ = 1.2 bar

T = 310 K

Since, the given piston has no mass and it is friction less. Hence, the system will be completely insulated in nature.

It is given that the state of surrounding is given as follows.

T = 298 K and P = 0.25 bar

So, $\Delta S_{surrounding}$ = 0

According to the first law of thermodynamics is as follows.

$\Delta U = Q + W$

Since, there is no friction so W = 0. Also, the system is well insulated hence, Q = 0. Therefore,

$\Delta U = Q + W$

= 0 + 0

= 0

For an ideal gas,

$\Delta U = C_{v} \times \Delta T$

As, $\Delta T$ = 0 this means that $T_{1} = T_{2}$

$P_{1}V_{1} = P_{2}V_{2}$

$\frac{V_{1}}{V_{2}} = \frac{P_{2}}{P_{1}}$

= $\frac{1.2}{6}$

= 0.2

Now, expression for entropy of the system is as follows.

$\Delta S_{system} = C_{p} ln \frac{T_{2}}{T_{1}} - R ln (\frac{P_{2}}{P_{1}})$

As, $T_{1} = T_{2}$

so, $\Delta S_{system} = -R ln (\frac{P_{2}}{P_{1}})$

= -R ln (0.2)

= $-8.314 \times ln(0.2)$

= $-8.314 \times -1.609$

= 13.38 mol K

Also, $\Delta S_{system} = \Delta S \times n$

= $13.38 \times 3$

= 40.142 J/K > 0

This means that $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$ > 0

= 40.142 J/K > 0

This means that the total entropy is immersible and follows second law of thermodynamics.

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3.00 mol of an ideal gas with CV, m=3R/2 undergoes an expansion from an initial state described by T...

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