Problem 2.3.2

 

A gas is compressed adiabatically from 1 atmosphere and 294.3 degrees Kelvin to a final pressure of 5 atmospheres.  Heat capacity at constant pressure is given as 7 calories per mole per degree Kelvin.  Heat capacity at constant volume is given as 5 calories per mole per degree Kelvin.  Gas law constant is 8.32 Pa. cubic meter per mole per degree Kelvin.  What is change in the internal energy, enthalpy and entropy?  How much work is done and how much heat is added or removed.  What are values of temperature and volume at the end of the operations?  Ratio of heat capacity at constant pressure and constant volume, k, is equal to 1.4. 

 

Solution: 

 

For an adiabatic process, when gas is changed from an initial condition of T1, P1 and V1 to a final condition of P2, then T2 can be calculated from T1 times (P2 over P1) raised to the power [(k-1)/k].  Final temperature is equal to 464.6 degree Kelvin.

 

Internal energy change, delta U, is equal to integral from T1 to T2 of CV dT.  For a constant value of CV, this simplifies to CV time (T2 minus T1).  Delta U is calculated as 3,542 joules per mole.  Note that 1 cal is equal to 4.184 joules. 

 

Enthalpy change, delta H, is equal to integral from T1 to T2 of CP dT.  For a constant value of CP, this simplifies to CP time (T2 minus T1).  Delta H is calculated as 4,958 joules per mole.

 

For an adiabatic process, heat added to or removed by the system is zero.  Entropy change for adiabatic process, reversible process is zero.

 

Applying first law of thermodynamics, work done by the system, W, is q minus delta U or minus 3542 joules per mole.  Minus sign indicates work is done on the system.

 

Initial volume, V1, of the gas can be found by RT1 over P1.  V1 is calculated to be 0.0237 cubic meters per mole.  Note that 1 atmosphere is equal to 1.01325 times 105 Pascals. 

 

Final volume, V2, of the gas can be found by V1 time (P1 over P2) to the power (1/k).  This is calculated to be 7.61 times 10-3 cubic meters per mole.