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Explain why the temperature of a gas increases when it is compressed adiabatically.

Short Answer

Expert verified

During compression, the work on the gas system and the change in internal energy are positive. As internal energy is a function of temperature, the temperature of the gas increases during adiabatic compression.

Step by step solution

01

Concept

By the first law of thermodynamics,\(\Delta U = Q - W\).

Now, for the adiabatic process, the change in heat energy is\(Q{\bf{ = 0}}\).

02

Explanation

For the adiabatic process,

\(\begin{aligned}{c}\Delta U = Q - W\\\Delta U = - W.\end{aligned}\)

During the adiabatic process, the change in the gas system's internal energy is equal to the opposite of the work done by the gas system or on the gas system.

During compression, the work on the gas system and the change in internal energy are positive. Internal energy is a function of temperature.

Hence, the temperature of the gas increases during adiabatic compression.

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Most popular questions from this chapter

(a) What happens if you remove the lid of a bottle containing chlorine gas? (b) Does the reverse process ever happen? Why or why not? (c) Can you think of two other examples of irreversibility?

Question: (II) (a) What is the coefficient of performance of an ideal heat pump that extracts heat from 6°C air outside and deposits heat inside a house at 24°C? (b) If this heat pump operates on 1200 W of electrical power, what is the maximum heat it can deliver into the house each hour? See Problem 35.

A 110 g insulated aluminum cup at 35°C is filled with 150 g of water at 45°C. After a few minutes, equilibrium is reached.

(a) Determine the final temperature and

(b) the total change in the entropy.

Question: (II) The pressure in an ideal gas is cut in half slowly, while being kept in a container with rigid walls. In the process, 465 kJ of heat left the gas. (a) How much work was done during this process? (b) What was the change in internal energy of the gas during this process?

Question: (III) The PV diagram in Fig. 15–23 shows two possible states of a system containing 1.75 moles of a monatomic ideal gas. \(\left( {{P_1} = {P_2} = {\bf{425}}\;{{\bf{N}} \mathord{\left/{\vphantom {{\bf{N}} {{{\bf{m}}^{\bf{2}}}}}} \right.} {{{\bf{m}}^{\bf{2}}}}},\;{V_1} = {\bf{2}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}},\;{V_2} = {\bf{8}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}}.} \right)\) (a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume \({V_2}\), followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of the gas for the two-step process B.

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