Answer :

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant=8.31Jmol^{-1}K^{-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 PV^{2}=constant.

So, differentiating this additional law as well we get

2PVdV + V^{2}dP=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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