One mole of ideal gas required 207 J heat to rise the temperature by 10°K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10°K the heat required is (R = 8/3 J/mole °K)(a) 1987 J(b) 29 J(c) 215.3 J(d) 124 J
sumit kumar , 10 Years ago
Grade Upto college level
3 Answers
Naveen Kumar
Last Activity: 10 Years ago
for 1 mole, Cp-Cv=R heat at constant pessure,Q=1*Cp*dT .. 207=Cp*10 Cp=20.7 heat at constant pressure, Q’=1*Cv*dT=(Cp-R)*10=(20.7-18.3)*10J=124J Hence answer is (D)
Siddharth Sharma
Last Activity: 6 Years ago
for 1 mole,Cp-Cv=Rheat at constant pessure,Q=1*Cp*dT. 207=Cp*10Cp=20.7heat at constant pressure,Q’=1*Cv*dT=(Cp-R)*10=(20.7-8.314)*10J=124JHence answer is (D)
Rishi Sharma
Last Activity: 4 Years ago
Dear Student, Please find below the solution to your problem.
for 1 mole, Cp-Cv=R heat at constant pessure, Q=1*Cp*dT 207=Cp*10 Cp =20.7 heat at constant pressure, Q’=1*Cv*dT =(Cp-R)*10 =(20.7-18.3)*10J =124J Thanks and Regards
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