Respuesta :
Answer:
Work done,[tex]w=5.12\times10^{6}\ \rm J[/tex]
change in internal Energy ,[tex]\Delta U=5.12\times10^6\ \rm J[/tex]
Explanation:
Given:
- Mass of helium gas [tex]m=3\ \rm kg[/tex]
- initial temperature [tex]T_i=286\ \rm K[/tex]
Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings
[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K[/tex]
Let n be the number of moles of Helium given by
[tex]n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3[/tex]
Work done in Adiabatic process
Let W be the work done
[tex]W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J[/tex]
The Internal Energy change in any Process is given by
Let [tex]\Delta U[/tex] be the change in internal Energy
[tex]\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J[/tex]
