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, $U=\frac{n f R T}{2}=\frac{5 R}{2}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$
Differentiate with respect to $t$
$\frac{d U}{d t}=\frac{5 R}{2}[\alpha+\beta t]$
But, $d Q=n C_{p} d T$
$ \therefore \frac{d Q}{d t} =n C_{p} \times \frac{d T}{d t} $
$\Rightarrow i^{2} r=\frac{7}{2} R \times[\alpha+\beta t] $
$ i =\sqrt{\frac{7 R}{2 r}(\alpha+\beta t)} $
$p V =n R T $
or $ V =\frac{n R T}{p}=\frac{n R}{p}\left(\alpha t+\frac{1}{2} \beta t^{2}\right) $
$X =\frac{n R}{p A}\left(\alpha t+\frac{1}{2} \beta t^{2}\right) $
$ v =\frac{n R}{p A}(\alpha+\beta t) $
and acceleration $=\frac{n R}{p A} \times \beta$
So, the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$ and the piston moves upwards with constant acceleration.