Question

An amount n (in moles) of a monatomic gas at an initial temperature T0 is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature Ts (> T0) and the atmospheric pressure is Pα. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness x and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time t.

Answer 1

The distance moved by the piston in time t is given by $d x=\frac{n r}{p o A}(T s-T o)\left(1-e^{-2 K \mathcal{A} t} / 5 x n R\right)$

Explanation:

Step 1:

Let us Consider,

Surface Area = A

Thickness = X

Thermal Conductivity = K

Atmospheric pressure = P

Mono atomic gas dQ = ncpd

$p d v=n R d T$

$p A d x=n R d T$

$d T=\frac{P A d x}{n r}$

Step 2:

The heat transfer is given by the following formula,

$\frac{d Q}{d t}=-\frac{K A}{x}(T s-T 0)$

On substituting the values of dQ, we get

$\frac{n c p d T}{d t}=\frac{K A}{x}(T s-T 0)$

$\frac{5}{2} n R \frac{d T}{d t}=\frac{K A}{x}(T s-T 0)$

$\frac{d T}{d t}=-\frac{2 K A}{5 x n R}(T s-T 0)$

Step 3:

On integrating the above equation with respect to time t,

$\int_{T s}^{T} \frac{d T}{(T s-T 0)}=\frac{-2 K A}{5 x n R} \int_{0}^{t} d t$

$\ln \frac{T s-T}{T s-T 0}=\frac{-2 K A t}{5 x n R}$

$T s-T=(T s-T 0) e^{-2 K A t} / 5 x n R$

$\begin{aligned}&T s-T=(T s-T 0) e^{-2 K A t} / 5 x n R\\&T=T s-(T s-T 0) e^{-2 K A t} / 5 x n R\end{aligned}$

$\Delta T=T-T 0$

Step 4:

On substituting the value of T in above equation, we get

$d T=T s-(T s-T 0) e^{-2 K A t} / 5 x n R^{-T 0}$

$d T=(T s-T 0)\left(1-e^{-2 K A t} / 5 x n R\right)$

As we know the value of dT,

$\frac{P A d x}{n R}=(T s-T 0)\left(1-e^{-2 K A t} / 5 x n R\right)$

$d x=\frac{n r}{p o A}(T s-T o)\left(1-e^{-2 K A t} / 5 x n R\right)$

Thus the distance moved by piston with respect to time period t is given by $d x=\frac{n r}{p o A}(T s-T o)\left(1-e^{-2 K \mathcal{A} t} / 5 x n R\right)$

Answer 2

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