Question

A heating element of resistance $r$ is fitted inside an adiabatic cylinder which carries a frictionless piston of mass $m$ and cross-section $A$ 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 $\Delta T=\alpha t+\frac{1}{2}\beta t^2(\alpha$ and $\beta$ are constants), while pressure remains constant. The atmospheric pressure above the piston is $P_0$. Then

• A. the rate of increase in internal energy is $\frac{5}{2}R\left(\alpha +\beta t\right)$
• B. the current flowing in the element is $\sqrt{\frac{5}{2r}R\left(\alpha +\beta t\right)}$
• C. the piston moves upwards with constant acceleration
• D. the piston moves upwards with constant speed

Answer 1

Answer: (A)

We know that
Internal energy, $U=\frac{nfRT}{2}=\frac{5R}{2}\left(\alpha t+\frac{1}{2}\beta t^2\right)$
Differentiate with respect to $t$
$\frac{dU}{dt}=\frac{5R}{2}[\alpha +\beta t]$
But, $dQ=n{C}_{p}dT$
$\therefore \frac{dQ}{dt}=n{C}_p\times \frac{dT}{dt}$
$\Rightarrow i^{2}r=\frac{7}{2}R\times [\alpha +\beta t]$
$i=\sqrt{\frac{7R}{2r}(\alpha +\beta t)}$
$pV=nRT$
or $V=\frac{nRT}{p}=\frac{nR}{p}\left(\alpha t+\frac{1}{2}\beta t^2\right)$
$X=\frac{nR}{pA}\left(\alpha t+\frac{1}{2}\beta t^2\right)$
$v=\frac{nR}{pA}(\alpha +\beta t)$
and acceleration $=\frac{nR}{pA}\times \beta$
So, the rate of increase in internal energy is $\frac{5}{2}R(\alpha +\beta t)$ and the piston moves upwards with constant acceleration.