Answered

When 19.3 J was added as heat to a particular ideal gas, the volume of the gas changed from 56.7 cm^3 to 104 cm^3 while the pressure remained constant at 0.947 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 1.97 x 10^-3 mol, find the molar specific heat of the gas at (b) constant pressure and (c) constant volume.

Answer :

Answer:

a)[tex] C_p=35.42\ \rm J/mol.K[/tex]

b)[tex] C_v=27.1\ \rm J/mol.K[/tex]

Explanation:

Given:

  • Heat given to the gas, [tex]Q=19.3\ \rm J[/tex]
  • Initial volume of the gas,[tex]V_i=56.7\ \rm cm^3[/tex]
  • Final volume of the gas, [tex]V_f=104\ \rm cm^3[/tex]
  • Constant pressure of the gas,[tex]P=0.947\ \rm atm[/tex]
  • Number of moles of the gas,[tex]n=1.97\times10^{-3}[/tex]

Let R be the gas constant which has value [tex]R=8.31432\times10^3\ \rm N\ m\ kmol^{-1}K^{-1}[/tex]

Work done in the process

[tex]W=PdV\\W=0.947\times(104-56.7)\times10^{-3}\ \rm L\ atm\\W=44.8\times10^{-3}\times101.33\ \rm J\\W=4.43\ \rm J[/tex]

Now Using First Law of thermodynamics

[tex]Q=W+\Delta U\\19.3=4.43+\Delta U\\\Delta U=14.77\ \rm J[/tex]

Let[tex]C_v[/tex] be the molar specific heat of the gas at constant volume given by

[tex]\Delta U=nC_v\Delta T\\14.77=\dfrac{C_v}{R}{PdV}\\14.77=\dfrac{C_v}{R}(0.987\times10^5(104-56.7)\times10^-6})\\C_v=3.26R\\C_v=27.1\ \rm J/mol.K[/tex]

Also we know that

Let[tex]C_p[/tex] be the molar specific heat of the gas at constant pressure given by

[tex]C_p-C_v=R\\C_p=R+C_v\\C_p=4.26R\\C_p=35.42\ \rm J/mol.K[/tex]

Other Questions