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3 years ago

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A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the

ideal gas. It is observed that the temperature of the ideal gas remains constant during this process as a result of heat transfer between the system and the surroundings at 30°C. Determine the entropy change of the ideal gas.

1 answer:

Sergeu [11.5K]3 years ago

8 0

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})$

Using the Boyle equation we have

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})$

Where,

$C_p$ = Specific heat at constant pressure

$T_1$ = Initial temperature of gas

$T_2$ = Final temprature of gas

R = Universal gas constant

$v_1$ = Initial specific Volume of gas

$v_2$ = Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

$T_1 = T_2, v_2=v_1$

The equation would turn out as

$\Delta S = C_p ln1-ln1$

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

$\Delta S = \frac{230kJ}{(30+273)K}$

$\Delta S = 0.76 kJ/K$

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

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