A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and crosssection A as shown in diagram. The cylinder contains one mole of an ideal diatomic gas. The current flows through the element ?

such that the temperature rises with time t asenter image source here

(C)(C). the piston moves up with constant acceleration .

(D)(D). the piston moves up with constant velocity.

1 Answer
Mar 16, 2018

A,CA,C

Explanation:

Well as the cylinder is adiabatic in nature,it will not release nor accept any heat other than the heat produced by the resistor.

Now.the pressure on the gas is solely due to the atmospheric pressure and weight of the piston,which don't change with the process.

So,it is an isobaric process.

Now,change in internal energy for an isobaric process is given as,

dU=nC_v dTdU=nCvdT

Now,C_v=f/2 RCv=f2R where,ff is the degrees of freedom,for diatomic gas it is 55

And,dT=alpha t +1/2 beta t^2dT=αt+12βt2

So,dU =1*5/2R(alphat +1/2 beta t^2)dU=152R(αt+12βt2)

so,(dU)/(dt)=5/2R(alpha +betat)dUdt=52R(α+βt)

So,option AA is correct.

Now,we know,power generated due to current flowing II through a ressitor rr is P=I^2rP=I2r

Now,here, P=(dQ)/(dt)P=dQdt (dQdQ is change in heat energy)

Now,for an isobaric process dQ=n C_pdT=7/2R(alphat+1/2betat^2)dQ=nCpdT=72R(αt+12βt2)

So,(dQ)/(dt) =7/2R(alpha+betat)=I^2rdQdt=72R(α+βt)=I2r

or,I=sqrt((7/(2r))R(alpha+betat))I=(72r)R(α+βt)

So,option BB is wrong.

Now,work done in isobaric process is dW =nRdT=P_odVdW=nRdT=PodV

now,P_o(dV)/(dt)=R(alpha +betat)PodVdt=R(α+βt)

As,V=AlV=Al (volume =area*length, given cross sectional area of piston is AA)

so,P_oA(dl)/(dt)=R(alpha+betat)PoAdldt=R(α+βt)

or,dl=R/(P_oA)(alphat +beta t)dtdl=RPoA(αt+βt)dt

or, int_0^ldl=R/(P_oA)int_0^t(alpha+betat)dtl0dl=RPoAt0(α+βt)dt

so,l=R/(P_oA)(alpha t +1/2 beta t^2)l=RPoA(αt+12βt2)

So,l prop t^2lt2

Compare with sprop t^2st2

So,the piston will move upward with constant acceleration.

So,option CC is correct as well