Five moles of neon gas at 2.00 atm and 27.0^∘ C is adiabatically compressed to one-third its ini

Five moles of neon gas at 2.00 atm and 27.0^∘ C is...

Key Concepts

Specific Heats and Gamma (?)

Specific heats at constant volume (cv) and constant pressure (cp) are intrinsic properties that determine how a gas stores energy. The ratio of these specific heats, known as gamma (? = cp/cv), plays a crucial role in adiabatic processes. It appears in the adiabatic condition equations (such as PV^? = constant and TV^(?-1) = constant), defining how pressure and temperature change with volume.

Thermodynamic Work in Adiabatic Processes

Work in thermodynamics during processes like adiabatic compression or expansion is defined as the energy transfer due to volume change under pressure. For an adiabatic process, since no heat is added or removed, the work done on or by the gas results in a direct change in its internal energy. Calculating this work often involves integrating pressure with respect to volume or using alternative energy balance methods.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged between the system and its surroundings. This means that any change in the internal energy of the gas is solely due to work done on or by the system. In adiabatic processes, the relationship between pressure, volume, and temperature is governed by specific adiabatic equations, reflecting the conservation of energy under these conditions.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates the pressure, volume, number of moles, and temperature of an ideal gas. This fundamental equation is used to determine one state variable when the others are known, and it provides the starting point for analyzing thermodynamic processes including compression and expansion.

Transcript

00:01 We have to find the value of first final pressure and final temperature and external work done on a gas and some values are given such that the solution for that five moles of neon gas is given that the moles of neon gas is equal to five mole and the temperature v1 is equal to v and the temperature v1 is equal to v and final temperature temperature v2 is equal to initial temperatures half and the initial pressure is equal to 2 and initial temperature is equal to 27 so we have to find the value of final pressure final temperature and work done and it is also given that the process is an adiatic process so we can write this the unit is 80 and two individuals and the value of in that 27 degrees celsius so for this we can calculate that first we know that the process is that adiatic so adiatic process delta q is equal to zero so now we can write for for adiavatic for adiatic adiatic p v -raged 2 power new is equal to constant or t we raise to power new minus 1 is equal to constant so new is equal to cp upon c v is equal to 1 .67 and c b is equal to 0 .148 calorie per gram degree celsius is that given and let final pressure is equal to p then p initial v initial raise to power that is new is equal to p final b final you and it is also showing called should be sometimes gamma so the value is 2 a tm is multiply by b raised 2 power 1 .67 is equal to p v rase to b upon 3 is raised to power 1 .67 that p is equal to 2 is multiply by 3 raised to power 1 .67 and this is the p final so the value of p final is equal to 12 .52680m in other form we can write this 12 .526 is 101 325 in pascal for pascal by converting so we get 1 .27 mega so p final is dead and the next one is how to calculate the value of t final.

03:39 This is the a part you can write this this is the final and the next one is we know that t1 v1 gamma or new minus 1 is equal to t 2 v2 new gamma minus 1 so by solving this 300 is multiply by v.

03:59 2 x2 power 1 .67 minus 1 is equal to t 2 v upon 3 raised to power 1 .67 minus 1 by solving this t 2 is equal to 300 is multiplied by 3 raise 2 power 1 point sorry this is 0 .67 so this is 0 .67 and this is 626 .34 so t 2 and t final is equal to 626 .314 kelvin in nearest we can write this t final and t2 is equal to 6 to 6 to 6 .6 calvin this is the second part and the third is as delta q is zero for an adiatic process for adiabatic so delta u is equal to delta w and delta u should be written as m c delta t so the value of m is equal to 5 is multiplied by 20 .18.

05:16 This is the value of m and c is equal to 3...

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