Question

One mole of an ideal monoatomic gas is kept in a rigid vessel. The vessel is kept inside a chamber whose temperature is ${87}^{\circ }\text{C}$. Initially, the temperature of the gas is $4^{\circ }\text{C}$. The walls of the vessel have an inner surface of area $500{\text{cm}}^2$ and thickness of $2\text{cm}$. If the temperature of the gas increases to $8^{\circ }\text{C}$ in $5.0\text{seconds}$, find the thermal conductivity of the material of the walls.

• A. $5\text{W/mK}$
• B. $0.05\text{W/mK}$
• C. $0.5\text{W/mK}$
• D. $50\text{W/mK}$

Answer 1

The correct option is B $0.05\text{W/mK}$

Initial temperature difference $\left(\Delta T_1\right)=87-4={83}^{\circ }\text{C}$
Final temperature difference $\left(\Delta T_2\right)=87-8={79}^{\circ }\text{C}$
As the change in temperature difference is small, we find
Average temperature difference $(\Delta T{)}_{avg}=\frac{\Delta T_1+\Delta T_2}{2}=\frac{83+79}{2}={81}^{\circ }\text{C}\text{or}(\Delta T{)}_{avg}=81\text{K}$
Rise in temperature of gas is $=8-4=4^{\circ }\text{C}\text{or}4\text{K}$
For monoatomic gas,
$C_V=\frac{3}{2}R=\frac{3}{2}\times 8.31=12.465\text{J/mol K}$
Since the walls are rigid, From first law of thermodynamics we can say that,
$\Delta Q=\Delta U$
So, heat supplied to the vessel containing gas in 5 seconds
$\Delta Q=n{C}_V\Delta T$
$=1\times 12.465\times 4$
$=49.86\approx 50\text{J}$
Let $K$ be the thermal conductivity of the wall.
Then, $\frac{\Delta Q}{\Delta t}=KA\frac{(T_1-T_2)}{L}$
$\Rightarrow \frac{50}{5}=\frac{K\times 500\times {10}^{-4}\times 81}{2\times {10}^{-2}}$
$[\because T_1-T_2=(\Delta T{)}_{avg}]$
$\Rightarrow K=\frac{10\times 2\times {10}^{-2}}{500\times {10}^{-4}\times 81}=0.04938\approx 0.05{\text{J s}}^{-1}{\text{m}}^{-1}{\text{K}}^{-1}$
i.e $K=0.05\text{W/mK}$