For the situation of the preceding problem, find (a) the increase in the gas temperature and (b)

For the situation of the preceding problem, find (a) the...

Key Concepts

First Law of Thermodynamics

This law states that the energy added to a system as heat minus the work done by the system equals the change in its internal energy. It is fundamental in determining how work and heat transfer contribute to temperature changes in a gas during processes such as compression or expansion.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the surroundings, meaning that any work done on or by the system directly affects its internal energy and temperature. Understanding adiabatic behavior is key when analyzing scenarios where temperature changes occur solely due to work being done on the gas.

Internal Energy of an Ideal Gas

In an ideal gas, the internal energy depends only on its temperature. This concept is essential for linking the work done on or by the gas to the resultant change in temperature, since any increase in internal energy (as measured by temperature rise) can be quantitatively evaluated using the specific heat capacity.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change a unit mass of a substance by one degree in temperature. In thermodynamic problems, it is used to calculate the heat removed or added to achieve a desired temperature change, such as returning the gas to its original temperature after an adiabatic process.

Transcript

00:01 Belongs to the laws of thermodynamics in which we have a balloon having volume v1 equals to 2 .5 liter and pressure p equals to 1 .0 atm and temperature t1 it is equals to 20 degree centigrade that is 293 kelvin okay so for the part a we have to determine the increase in the temperature when the volume of the bubble volume of the balloon so v2 this has been dropped by the 10 % so v2 will be equals to 0 .9 times of v1.

00:33 So from here we have given that delta v it is equals to 10 % of initial volume v1.

00:41 So from the at the constant pressure we can write from the ideal gas that volume is directly proportional to temperature t so v2 by v1 this will be equals to t 2 by t1.

00:51 So from here we get t2 that is equals to v2 which is 0 .9 times of v1 divided by v1, but be temperature t1 which is 293 kelvin so 293 kelvin okay so from here we get temperature t2 this is equals to temperature t2 this will be equals to a 0 .9 multiplied by 2 793 kelvin okay so we have to determine the change in temperature so delta t it will be equals to t2 minus t1 so we get 0 .9 multiplied by 293 minus 293 kelvin so from here we get change in temperature delta t it is equals to minus 29 .3 calvin so this become the answer for this question this is the change in temperature when the volume of the bubble has been decreased or dropped by 10 % okay now for the second part b we have to determine the work done so work done w it will be given by pressure multiplier by change in volume okay so substituting the values so we get minus also so, so minus pressure is 1 atm.

02:00 So 1 atm is 1 .1 .3 kilo pascal that is 10 to the power 3 pascal and delta v which is v2 minus v1.

02:08 So v2 is 0 .9 v1 minus v1...

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