Question

The centre of the sun consists of gases, whose average molecular weight is 2. If the density of the gases is $2.73\times {10}^{3}kg/m^3$, at a pressure of $1.12\times {10}^9$atm, the temperature at the centre of the sun is? (Assuming ideal behaviour of gases)
A) ${10}^{8}K$
B) ${10}^6°C$
C) ${10}^{7}K$
D) ${10}^{9}K$

Answer 1

Hint: As per the given question the conditions are ideal then we will use ideal gas equation in order to bring out the temperature at the centre of the sun;
Equation of ideal gas is given as;
$PV=\mu RT$ (P is the pressure, V is the volume, R is the gas constant and T is the temperature)
$\mu$ is ${\displaystyle \frac{m}{M}}$, M is the average molecular weight of the sun, m is the mass of gases.

Complete step by step solution:
Let’s discuss in more detail about the ideal gas equation.
We have three gas laws: Boyle’s law (law of constant temperature) Charles’ law (law of constant volume) and Gay Lussa’s law (law of constant pressure), when the simultaneous change in volume, pressure and temperature takes place gas equation is formed stated below:
$PV=\mu RT$
In which R is the gas constant having a constant value $0.0831LatmMo{l}^{-1}{K}^{-1}$.
The gas which obeys all the gas laws is called an ideal gas.
Now we will come to the calculation part of the question:
$\Rightarrow PV=\mu RT$.....................(1)
We can write $\mu$ as ${\displaystyle \frac{m}{M}}$
Where, m=$\rho V$ therefore we can write $\mu$as ${\displaystyle \frac{\rho V}{M}}$
Substituting the value of $\mu$ in equation 1
$\Rightarrow PV={\displaystyle \frac{\rho V}{M}}RT$
On cancelling V on LHS and RHS
$\Rightarrow P={\displaystyle \frac{\rho }{M}}RT$ ............(2)
From equation 2 we will arrange the terms to get the value of T
$$\begin{gathered} \Rightarrow {\displaystyle \frac{PM}{\rho R}}=T \\ \Rightarrow T={\displaystyle \frac{PM}{\rho R}} \end{gathered}$$.........................(3)
In equation 3 we will substitute all the numerical values.
$$\begin{gathered} \Rightarrow T={\displaystyle \frac{1.12\times {10}^9\times 2}{2.73\times {10}^3\times 0.0831}} \\ \Rightarrow T={\displaystyle \frac{2.24\times {10}^9}{22.43\times {10}^3}} \end{gathered}$$ (Simple multiplication is done)
$\Rightarrow T=9.98\times {10}^6$
We assume 9.98 as 10 therefore our solution will become; $T={10}^{7}K.$

Option (C) is correct.

Note: Ideal gas equation has many relations embedded in it, which has different applications like density of gas is directly proportional to the pressure and molecular mass and inversely proportional to the temperature, this relation is used for extinguishing fire by spreading $C{O}_2$ carbon dioxide over the fire because density of $C{O}_2$ is more than oxygen and thus carbon dioxide acts as blanket over the fire in order to reduce the presence the oxygen (because burning needs oxygen).