One mole of an ideal monatomic gas (Cv,m = 3/2 R) at 320 K isallowed to expand a
ID: 690600 • Letter: O
Question
One mole of an ideal monatomic gas (Cv,m = 3/2 R) at 320 K isallowed to expand adiabatically reversyble against a constantexternal pressure of 1.98 atm from a volume of 6.0 L to 14.0L. Calculate q, w, DeltaU, and DeltaH for the process. Note: system is adiabatic reversible! One mole of an ideal monatomic gas (Cv,m = 3/2 R) at 320 K isallowed to expand adiabatically reversyble against a constantexternal pressure of 1.98 atm from a volume of 6.0 L to 14.0L. Calculate q, w, DeltaU, and DeltaH for the process. Note: system is adiabatic reversible!Explanation / Answer
We know that for an adiabatic process q = 0 For adiabatic process TV ( - 1 ) =constant TV ( - 1 ) = T'V' ( - 1) For mono atomic gases = 1.66 TV ( - 1 ) = T'V' ( - 1) T' / T = V ( - 1 ) / V' ( - 1 ) = (V / V' )( - 1 ) = ( 6 /14 ) ( 1.66 - 1 ) =0.5716 T' = 0.5716 * T = 0.5716 * 320 K = 182.93 K ~ 183 K Work done in an adibatic process , W = [ nR / (-1) ] ( T- T' ) = [ 1* 8.314 / (1.66-1) ] ( 320-183) = 1725.785 J So work done W = 1725.785 J We know that U = q - W = 0 -W For adiabatic process q = 0 =-1725.785 J = ( 6 /14 ) ( 1.66 - 1 ) =0.5716 T' = 0.5716 * T = 0.5716 * 320 K = 182.93 K ~ 183 K Work done in an adibatic process , W = [ nR / (-1) ] ( T- T' ) = [ 1* 8.314 / (1.66-1) ] ( 320-183) = 1725.785 J So work done W = 1725.785 J We know that U = q - W = 0 -W For adiabatic process q = 0 =-1725.785 J = 0 -W For adiabatic process q = 0 =-1725.785 J We know that H = U + PV = U + W = -1725.785 J + -1725.785 J = 0Related Questions
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