# 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 as $\left(C\right)$. the piston moves up with constant acceleration . $\left(D\right)$. the piston moves up with constant velocity.

Mar 16, 2018

$A , C$

#### 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,

$\mathrm{dU} = n {C}_{v} \mathrm{dT}$

Now,${C}_{v} = \frac{f}{2} R$ where,$f$ is the degrees of freedom,for diatomic gas it is $5$

And,$\mathrm{dT} = \alpha t + \frac{1}{2} \beta {t}^{2}$

So,$\mathrm{dU} = 1 \cdot \frac{5}{2} R \left(\alpha t + \frac{1}{2} \beta {t}^{2}\right)$

so,$\frac{\mathrm{dU}}{\mathrm{dt}} = \frac{5}{2} R \left(\alpha + \beta t\right)$

So,option $A$ is correct.

Now,we know,power generated due to current flowing $I$ through a ressitor $r$ is $P = {I}^{2} r$

Now,here, $P = \frac{\mathrm{dQ}}{\mathrm{dt}}$ ($\mathrm{dQ}$ is change in heat energy)

Now,for an isobaric process $\mathrm{dQ} = n {C}_{p} \mathrm{dT} = \frac{7}{2} R \left(\alpha t + \frac{1}{2} \beta {t}^{2}\right)$

So,$\frac{\mathrm{dQ}}{\mathrm{dt}} = \frac{7}{2} R \left(\alpha + \beta t\right) = {I}^{2} r$

or,$I = \sqrt{\left(\frac{7}{2 r}\right) R \left(\alpha + \beta t\right)}$

So,option $B$ is wrong.

Now,work done in isobaric process is $\mathrm{dW} = n R \mathrm{dT} = {P}_{o} \mathrm{dV}$

now,${P}_{o} \frac{\mathrm{dV}}{\mathrm{dt}} = R \left(\alpha + \beta t\right)$

As,$V = A l$ (volume =area*length, given cross sectional area of piston is $A$)

so,${P}_{o} A \frac{\mathrm{dl}}{\mathrm{dt}} = R \left(\alpha + \beta t\right)$

or,$\mathrm{dl} = \frac{R}{{P}_{o} A} \left(\alpha t + \beta t\right) \mathrm{dt}$

or, ${\int}_{0}^{l} \mathrm{dl} = \frac{R}{{P}_{o} A} {\int}_{0}^{t} \left(\alpha + \beta t\right) \mathrm{dt}$

so,$l = \frac{R}{{P}_{o} A} \left(\alpha t + \frac{1}{2} \beta {t}^{2}\right)$

So,$l \propto {t}^{2}$

Compare with $s \propto {t}^{2}$

So,the piston will move upward with constant acceleration.

So,option $C$ is correct as well