Question 104: Thermodynamics & KTG

Question

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of $\Delta Q : \Delta U : \Delta W$ is ______.

Accepted Answer

The diameter of molecule is, $d=5 	imes 10^{-10} \mathrm{~m}$
Temperature, $\mathrm{T} 41^{\circ} \mathrm{C}=(41+273) \mathrm{K}=314 \mathrm{~K}$
Pressure, $\mathrm{P}=1.38 	imes 10^5 \mathrm{~Pa}$
Boltzmann constant, $\mathrm{k}_{\mathrm{B}}=1.38 	imes 10^{-23} \mathrm{~J} / \mathrm{K}$
The standard formula for the mean free path $\lambda$ is:
$$ \lambda=\frac{\mathrm{k}_{\mathrm{B}} \mathrm{~T}}{\sqrt{2} \pi \mathrm{~d}^2 \mathrm{P}} $$
Substituting the values,
$$ \lambda=\frac{\left(1.38 	imes 10^{-23}ight) 	imes 314}{\sqrt{2} 	imes \pi 	imes\left(5 	imes 10^{-10}ight)^2 	imes\left(1.38 	imes 10^5ight)} $$
$$\Rightarrow $$ $\lambda=\frac{10^{-23} 	imes 314}{\sqrt{2} 	imes \pi 	imes 25 	imes 10^{-20} 	imes 10^5}$
$$\Rightarrow $$ $\lambda=\frac{314}{\sqrt{2} 	imes \pi 	imes 25 	imes 10^{-8}}$
Using $\pi \approx 3.14$ :
$$ \lambda=\frac{314}{\sqrt{2} 	imes 3.14 	imes 25 	imes 10^{-8}} $$
$$\Rightarrow $$ $\lambda=\frac{4}{\sqrt{2}} 	imes 10^{-8}=2 \sqrt{2} 	imes 10^{-8} \mathrm{~m}$
Hence, the correct option is D.

Answer

2 : 3 : 5

Answer

5 : 3 : 2

Answer

2 : 5 : 7

Answer

7 : 5 : 2

Thermodynamics & KTG

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of \Delta Q : \Delta U : \Delta W is ______.

Choose the correct answer: