1.

A heating element of piston 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 as DeltaT =alpah t+1/2beta t^2 ? (alphaand betaare constants), while pressure remains constant. The atmospheric pressure above the piston is P_0. Then

Answer»

the RATE of increase in internal energy is `5/2 R ( alpha + betat )`
the current FLOWING in the element is `sqrt((5)/( 2r ) R(alpha+ betat ))`
the piston moves upwards with CONSTANT acceleration,
the piston moves upwards with constant speed.

Solution :increase ininternalenergy ` dU = n C_V d T `
` therefore` Rateof increase in INTERNALENERGY
` (dU)/( d t)= n C_V (d T)/(d t)`
here ` n=1, dT=alpha t+1/2betat^2therefore(dT)/( d t)= alpha+ betat`
fordiatomicgas ` C_V=5/2R ,C_P= 7/2R `
`therefore(dU )/( d t)= 1 xx5/2R xx( alpha+ betat)= 5/2R ( alpha+ betat )`
heatsuppliedby theheatingelement to thegas atconstantpressure
` d Q= n C_(P )d T `
Rateof hetasupplied
` ( dQ)/(d t)= nC_P( d t)/( d t) = 1 xx (7)/(2)R xx ( alpha+ beta t )` ( using (i ))
` =7/2R ( alpha+ beta t )`
LetI becurrentflowingin theelement
` thereforeI^2r =7/2R( alpha+ betat )`
` I^2 = 7/( 2 r)R (alpha + betay)impliesI= sqrt( 7/(2 r)R (alpha+ betat))`
accordingto idealgas equation`PV = nRT `
At contain pressure`PdVn RT `
` PdV= nR( alphat + 1/2beta t^2 )`
workdone` w =pdV `
` =n R( alpha t + 1/2beta t^2 )`
LetF beforceexertedbythegasonthe pistonof areaA
` thereforeP = ( F)/( A)orF = pA `
letthepistonmovedupwardsby adistancex .
` thereforeW = F xorx =W /F= (nR)/(pA )( alphat +1/2beta t ^2)`( using (ii ) )
veloity` v=(dx )/ (dt)=(nR)/( PA)( alpha+ betat )`
Acceleration` a = ( dv )/( dt ) = ( nR) /(PA )beta `


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