Question

One mole of an ideal monatomic gas is kept in a rigid vessel. The vessel is kept inside a steam chamber whose temperature is ${97}^{\circ }\text{C}$. Initially, the temperature of the gas is $5^{\circ }\text{C}$. The walls of the vessel have an inner surface of area $800{\text{cm}}^2$ and thickness $1\text{cm}$. If the temperature of the gas increases to $9^{\circ }\text{C}$ in $5$ seconds, then:

• A. Heat supplied to the gas is $4.98\text{J}$.
• B. Heat absorbed from the gas is $49.8\text{J}$.
• C. Thermal conductivity of the material of the walls is $0.014\text{J}{\text{s}}^{-1}{\text{m}}^{-1}{\text{K}}^{-1}$.
• D. Thermal conducitivity of the material of the walls is $0.14\text{J}{\text{s}}^{-1}{\text{m}}^{-1}{\text{K}}^{-1}$.

Answer 1

The correct option is C Thermal conductivity of the material of the walls is $0.014\text{J}{\text{s}}^{-1}{\text{m}}^{-1}{\text{K}}^{-1}$.

The initial temperature difference is ${97}^{\circ }\text{C}-5^{\circ }\text{C}={92}^{\circ }\text{C}=92\text{K}$

At $5\text{s}$, the temperature difference becomes ${97}^{\circ }\text{C}-9^{\circ }\text{C}={88}^{\circ }\text{C}=88\text{K}$

As the change in temperature difference is small, we will work with the average temperture difference
$\Delta T=\frac{92+88}{2}=90\text{K}$

Also, thickness of the wall, $L=1\text{cm}={10}^{-2}\text{m}$

The rise in temperature of the gas is
$\Delta T_g=9^{\circ }\text{C}-5^{\circ }\text{C}=4^{\circ }\text{C}=4\text{K}$

Now, heat supplied to the gas (as temperature increases) in $5\text{s}$ is
$\Delta Q=n{C}_v\Delta T_g$
$=1\times \frac{3}{2}\times 8.3\times 4=49.8\text{J}$
$[\because C_v=\frac{3R}{2}\text{for monoatomic gas}]$

If the thermal conductivity is $K$, then
$\frac{49.8}{5}=\frac{K\times (800\times {10}^{-4}\times 90)}{1\times {10}^{-2}}$
$[\because \frac{dQ}{dt}=\frac{KA\Delta T}{L}]$
$\Rightarrow K=\frac{49.8}{3600}=0.014\text{J}{\text{s}}^{-1}{\text{m}}^{-1}{\text{K}}^{-1}$

Therefore, only option (c) is correct.