One mole of an ideal gas requires $$207$$ J heat to raise it’s temperature by $$10$$ K when heated at constant pressure. If the same gas is heated at constant volume the heat required to raise the temperature by $$10$$ K for the same gas, will be : (given$$ $$R=8.3$$ $$J/mol/K)

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Infinity Learn · Accepted Answer

amount of heat required to raise the temperature of gas at constant pressure is $$Qp=nCp$$∆Tgiven $$Qp=207Joule$$;number of mole $$n=1$$;change in temperature ∆$$T=10K$$     substitute the given values in above equation of Qp  $$207=1$$×Cp×$$10$$.........(1)amount$$ of heat required to raise the temperature of gas at constant volume is $$Qv=nCv$$∆Tgiven $$n=1$$;∆$$T=10K$$, substitute in Qv  $$Qv=1$$×Cv×$$10$$ ........$$(2)  subtract, $$eqn(1)-eqn(2)$$,  (207-Qv)=10(Cp$$−$$Cv) ⇒$$207-Qv=10$$×$$8.3=83$$ ⇒$$Qv=207-83$$ ⇒$$Qv=124$$ J

One mole of an ideal gas requires 207 J heat to raise it’s temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume the heat required to raise the temperature by 10 K for the same gas, will be : (given R=8.3 J/mol/K)

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