Thermodynamics Test

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A process in which no heat enters or leaves a system is called an adiabatic process, $$Q=0$$. This prevention of heat flow may be accomplished either by surrounding the system with a thick layer of heat- insulating material(such as cork, asbestos, fire brick or styrofoam), or by performing the process quickly. The flow of heat requires finite time. So any process performed quickly enough will be practically adiabatic. Applying the first law to an adiabatic process, we get.

$$U_2-U_1=\Delta U=-W$$(adiabatic process)
Thus, the change in the internal energy of a system, an adiabatic process is equal in magnitude to the work done by the system. If the work $$W$$ is negative, as when a system is compressed, then $$W$$ is positive, $$U_2$$ is greater than $$U_1$$ and the internal energy of the system increases. If $$W$$ is positive, as when system expands, the internal energy of the system decreases. An increase of internal energy is usually(but not always) accompanied by a rise in temperature, and a decrease in internal energy by a temperature drop.

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One mole of an ideal gas is taken from an initial state $$P_o, V_o$$ and temperature $$T_o$$ through the following activities. ($$\gamma = \dfrac{C_p}{C_v}, C_p = \dfrac{7R}{2}, C_v = \dfrac{5R}{2}$$)
(i) Heating at constant volume to a temperature three times.
(ii) Adiabatic expansion to a volume $$2V_o$$
(iii) Cooling at constant volume to a temperature on third
(iv) Adiabatic compression so that it is returned to its initial state

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The work done during adiabatic reversible process is given by $$\displaystyle w=\dfrac { nR[{ T }_{ f }-{ T }_{ i }] }{ \gamma -1 } $$. Thus, if $$\displaystyle { T }_{ f }>{ T }_{ i }$$, then since $$\displaystyle \gamma-1=+ve\ (\because \gamma >1)$$, the work done is +ve i.e., work is done on gas. Also if $$\displaystyle { T }_{ f }<{ T }_{ i },w=-ve$$, thus work is done by the gas.

The adiabatic process are more steeper than isothermal process and slope of adiabatic process $$\displaystyle =\gamma \times $$slope of isothermal process and slope of adiabatic process > slope of isothermal process (since $$\displaystyle \gamma >1$$). Also the adiabatic process obey $$\displaystyle { PV }^{ \gamma }$$= constant, whereas in isothermal process PV = constant.

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One mole of a monoatomic ideal gas is taken through the cycle ABCDA as shown in the figure. $$T_A\, =\, 1000\, K$$ and $$2P_A\, =\, 3P_B\, =\, 6P_C$$.
$$\left [Assume\quad \left (\displaystyle \frac{2}{3}\right )^{0.4}\, =\, 0.85\, \ and\ \, R\, =\, \displaystyle \frac{25}{3}\, JK^{-1}\, mol^{-1}\right ]$$

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