Derive an expression for the work done by a gas undergoing expansion from volume V1 to V2.

Let us consider an ideal gas enclosed in a cylinder fitted with massless and frictionless piston. Let A be the area of cross-section of piston. Let V be the volume and P be the pressure exerted by gas on the piston. The piston is kept in equilibrium by applying pressure P from outside.

Let us consider an ideal gas enclosed in a cylinder fitted with massl

Let the applied pressure be decreased by infinitesimally small amount, so that the piston moves by infinitesimal distance dx. The amount of work done by the gas in infinitesimal expansion is,
dW space equals space straight F. dx space equals space straight P space straight A space dx
space space space space space space space space equals space straight P space dV space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis               
where dV = Adx, the infinitesimal increase in volume. Total work done by gas in expanding it from volume V1 to V2 can be obtained by integrating equation (1) from V1 to V2 i.e.
straight W equals integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript PdV


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Derive the expression for the work done by the gas during isothermal expansion.


We know work done by or on the system is given by

We know work done by or on the system is given by             
                          

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Derive an expression for the work done in adiabatic process.

Consider a unit mole of gas contained in a perfectly non-conducting cylinder provided with a non-conducting and frictionless piston.

Let C
v be the specific heat of gas at constant volume.

Let at any instant, when the pressure of gas is P, the gas be compressed by small volume dV.

Then work done on the gas is, 

                          dW = PdV 

Total work done on gas to compress from volume V1 to V2 is given by 

     integral dW space equals space straight W space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript straight P space dV                   ...(1) 

According to first law of thermodynamics, 

              dQ space equals space dU plus PdV 

For adiabatic process,  dQ space equals space 0 

∴           PdV = -dU = negative straight C subscript straight v dT 

∴   straight W space equals space integral subscript straight V subscript 1 end subscript superscript straight V subscript 2 end superscript PdV space equals space integral subscript straight T subscript 1 end subscript superscript straight T subscript 2 end superscript minus straight C subscript straight v dT 

where,

T
1 is the temperature of gas when volume is V1 and T2 when volume is V2

Thus, work done is given by, 

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        equals negative straight C subscript straight v left parenthesis straight T subscript 2 minus straight T subscript 1 right parenthesis space equals space straight C subscript straight v left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis

        equals fraction numerator straight R over denominator straight gamma minus 1 end fraction left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis space equals space fraction numerator 1 over denominator straight gamma minus 1 end fraction left parenthesis RT subscript 1 minus RT subscript 2 right parenthesis
equals space fraction numerator 1 over denominator straight gamma minus 1 end fraction left parenthesis straight P subscript 1 straight V subscript 1 minus straight P subscript 2 straight V subscript 2 right parenthesis 

Therefore,

space straight W subscript adi space equals space straight C subscript straight v left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis space equals space fraction numerator straight R over denominator straight gamma minus 1 end fraction left parenthesis straight T subscript 1 minus straight T subscript 2 right parenthesis
space space space space space space space space space space space equals space fraction numerator 1 over denominator straight gamma minus 1 end fraction left parenthesis straight P subscript 1 straight V subscript 1 minus straight P subscript 2 straight V subscript 2 right parenthesis 

The above expression gives us the amount of work done in adiabatic process. 
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Derive the equation of state for adiabatic process. Plot P versus V graph for the process.

According to the first law of thermodynamics,
dQ space equals space dU plus dW
In adiabatic process, no heat is allowed to exchange between the system and surrounding, therefore dQ = 0. Thus,

According to the first law of thermodynamics,In adiabatic process, no

                 
   
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What is the equation of state for isothermal process? Plot P versus V graph and P versus T graph for isothermal process for an ideal gas. Discuss the isothermal process using first law of thermodynamics. 

An ideal gas equation is,
PV = nRT
Since in isothermal process, T is constant, therefore the equation of state for isothermal process reduces to
PV = constant
P versus V graph and P versus T graph for isothermal process is as shown below.

An ideal gas equation is,PV = nRTSince in isothermal process, T is co
According to first law of thermodynamics,

An ideal gas equation is,PV = nRTSince in isothermal process, T is co


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