Question

how to find moles when Cp=58 and Cv=24?​

Answer 1

Given that:-

Given that:-P

Given that:-P ext.

Given that:-P ext.

Given that:-P ext. =2atm

Given that:-P ext. =2atmP

Given that:-P ext. =2atmP 1

Given that:-P ext. =2atmP 1

Given that:-P ext. =2atmP 1 =5atm

Given that:-P ext. =2atmP 1 =5atmP

Given that:-P ext. =2atmP 1 =5atmP 2

Given that:-P ext. =2atmP 1 =5atmP 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atm

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300K

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V =

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 R

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V 2

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V 2 −V

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V 2 −V 1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V 2 −V 1

Given that:-P ext. =2atmP 1 =5atmP 2 =2atmT 1 =300KT 2 =T(say)=?n=1C V = 25 RFor an adiabatic process,W=nC V dT=−PΔV∴nC V (T 2 −T 1 )=−P ext (V 2 −V 1 )