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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
anser 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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