2.20 The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 10⁵ cm³ $\approx 27\times 10^{4}c{m}^3$  

$K=\frac{\left(\frac{Q}{r}\right)\mathrm{\Delta }x}{\mathrm{\Delta }AT}$

$\begin{array}{ll}\Rightarrow & \frac{\left({\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}\right)\left(\mathrm{L}\right)}{\left({\mathrm{L}}^2\right)\left(\theta \right)\left(\mathrm{T}\right)}\\ \Rightarrow \mathrm{}& {\mathrm{M}}^1{\mathrm{L}}^{1-}{\mathrm{T}}^{-3}{\theta }^{-1}\end{array}$

Viscosity = Pascal . Second

$P^X{A}^Y{T}^Z=\left[M^1{L}^{-1}{T}^{-1}\right]$        

=  ${\left[M^1{L}^{+1}{T}^{-1}\right]}^x\left[L^2\right]y{\left[T^1\right]}^z=M^{-1}{L}^{-1}{T}^{-1}$

= x = 1,     x + 2y = -1,      -x + z = 1

y = -1,              Z = 0

Viscosity = [P¹A^-1T⁰]

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10^-2 cm

$\Delta Q=ms\Delta T\Rightarrow s=\frac{\Delta Q}{m\Delta T}\Rightarrow \left[s\right]=\frac{M{L}^2{T}^{-2}}{MQ}=L^2{T}^{-2}{Q}^{-1}$

$\Delta Q=mL\Rightarrow \left[L\right]=\frac{M{L}^2{T}^{-2}}{M}=L^2{T}^{-2}$