Table of Contents
Why does an ideal gas expand in vacuum w 0?
Since the gas is expanding into a vacuum, pext=0 and dw=−pextdV=0. By the First Law, dU=dq+dw=0 and temperature is therefore constant (since the temperature of an ideal gas depends fully on U).
Is the isothermal expansion of a gas into a vacuum spontaneous?
During this constant temperature, or isothermal, expansion into a vacuum, the gas does no work (w = 0) and no heat is transferred between the system and surroundings (q = 0). Thus, E = 0 for the expansion. Nevertheless, the process is spontaneous.
Is Q 0 in free expansion?
When there will be no work done, there is no question of heat transfer, as the heat transfer only takes place where there is work done. Thus, the term Q also equals to 0. Thus, the change in the internal energy of the ideal gas in free expansion is zero.
What is Q for ideal gas?
Heat Capacity at Constant Volume For an ideal gas, applying the First Law of Thermodynamics tells us that heat is also equal to: Q = ΔEint + W, although W = 0 at constant volume. This is from the extra 2 or 3 contributions to the internal energy from rotations.
Why does an ideal gas expand in a vacuum?
Since, vacuum is just empty space, it has no surrounding. Therefore, no heat is exchanged by the gas.
What happens when an ideal gas expands in vacuum?
When an ideal gas expands into the vacuum it does zero work. Since the gas does not have to expend any internal energy for the expansion, the term free expansion. Since there is no change in internal energy, the temperature of the ideal gas does not change on free expansion.
How does the answer to part B explain the spontaneous expansion of the gas?
Such a reverse reaction is non spontaneous. A process that is spontaneous in one direction is non spontaneous in the opposite direction. Consider the following physical and chemical processes as a way of looking and identifying spontaneous processes.
What happens to the temperature of an ideal gas that is compressed adiabatically?
When an ideal gas is compressed adiabatically (Q=0), work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops. In fact, the temperature increases can be so large that the mixture can explode without the addition of a spark.
Which is zero for ideal gas in isothermal free expansion?
To be more precise, isothermal expansion gives ∆T = 0 (no change in the temperature). When the vacuum gets expanded, it leads to the free expansion of a gas. In the case of an ideal gas, the rate of free expansion is NIL, that is, the work done is 0.
What is isothermal expansion of an ideal gas?
Hint: The isothermal expansion means a gas is expanding from initial volume to final volume at constant temperature. The internal energy and enthalpy depend upon the temperature only in case of ideal gas. So, during the isothermal expansion of an ideal gas, the temperature remains constant and volume increases.
What is the work done during free expansion of ideal gas?
During free expansion of an ideal gas, the work done is 0 be it a reversible or irreversible process. It is known that the change in internal energy of a system is given as:
What is the work done when an ideal gas is isothermal?
When an ideal gas is subjected to isothermal expansion (∆T = 0) in vacuum the work done w = 0 as p ex =0. As determined by Joule experimentally q =0, thus ∆U = 0. For isothermal reversible and irreversible changes; equation 1 can be expressed as:
How do you find the work done by isothermal expansion?
When an ideal gas is subjected to isothermal expansion (∆T = 0) in vacuum the work done w = 0 as p ex =0. As determined by Joule experimentally q =0, thus ∆U = 0. For isothermal reversible and irreversible changes; equation 1 can be expressed as: Isothermal reversible change: q = -w = p ex (V f -V
What is the internal energy of an ideal gas?
Internal energy of ideal gas depends on temperature only. When a gas expand in vacuum it’s called free expansion. Since there’s nothing to oppose the gas, work done is 0. Also, in free expansion there’s no heat loss or gain. Now, according to 1st law of thermodynamics, Since work done is 0 and no heat loss/gain, dU = 0.