Answer :

We know ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant=8.31Jmol-1K-1


T=temperature


n=number of moles of gas


P=pressure of gas


So ………. (I)


Now differentiating the ideal gas equation, we get


PdV + VdP=nRdT …………. (II) (we have applied product rule for


differentiation of PV)


Now as given in question the ideal here follows and additional law which is PV2=constant.


So, differentiating this additional law as well we get


2PVdV + V2dP=0


Taking V as common we get


2PdV + VdP=0 ……… (III)


Subtract equation (III) from (II)


2PdV + VdP - PdV - VdP = -nRdT


PdV=-nRdT


From equation (I), substitute the value of P in above equation we get




Integrating equation (IV) from limits V to 2V and T1 to T2



We know . Applying this formula




Where we have applied the property of ln which is


ln(a)-ln(b)=ln(a/b)




So, the temperature at which the gas expands is half of the initially temperature.


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