The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K is:Take R =2 cal / K . mol A. 720 calB. 800 calC. 550 calD. 360 cal

您所在的位置：网站首页 › j·mol-1·k-1 › The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K is:Take R =2 cal / K . mol A. 720 calB. 800 calC. 550 calD. 360 cal

The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K is:Take R =2 cal / K . mol A. 720 calB. 800 calC. 550 calD. 360 cal

#The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K is:Take R =2 cal / K . mol A. 720 calB. 800 calC. 550 calD. 360 cal| 来源: 网络整理| 查看: 265

The correct option is A 720 calGiven: n=2 molPext=2 atm=P2P1=1 atmT=300K, R=2 cal/K.mol

for monoatomic gas, Cv=32R for adiabatic process q=0 ∴from Ist lawdU=w so, nCvdT=−PextdV nCv(T2−T1)=−Pext(V2−V1) nCv(T2−T1)=−Pext(nRT2P2−nRT1P1) n32R(T2−300)=−2(nRT22−nR×3001)

32(T2−300)=−T2+600 T2=420 K

so, work done = nCv(T2−T1) =2×32×2×(420−300) w=720 cal

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The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K is:Take R =2 cal / K . mol A. 720 calB. 800 calC. 550 calD. 360 cal

The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K is:Take R =2 cal / K . mol A. 720 calB. 800 calC. 550 calD. 360 cal

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