If the density of the Sun is 1.4 , g , cm^{-3 | vedclass

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(D) Using the ideal gas equation $PV = \mu RT$, where $\mu = \frac{m}{M_w}$ and $\rho = \frac{m}{V}$, we have $m = \rho V$. Substituting $\mu = \frac{

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$2.4 	imes 10^7 \, K$

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(D) Using the ideal gas equation $PV = \mu RT$, where $\mu = \frac{m}{M_w}$ and $ho = \frac{m}{V}$, we have $m = ho V$. Substituting $\mu = \frac{ho V}{M_w}$ into the ideal gas equation:$PV = \left( \frac{ho V}{M_w} ight) RT \Rightarrow T = \frac{P M_w}{ho R}$Given:$P = 1.4 	imes 10^9 \, 	ext{atm} = 1.4 	imes 10^9 	imes 1.013 	imes 10^5 \, N \, m^{-2} \approx 1.418 	imes 10^{14} \, Pa$$ho = 1.4 \, g \, cm^{-3} = 1400 \, kg \, m^{-3}$$M_w = 2 \, g \, mol^{-1} = 2 	imes 10^{-3} \, kg \, mol^{-1}$$R = 8.4 \, J \, mol^{-1} \, K^{-1}$Substituting the values:$T = \frac{(1.4 	imes 10^9 	imes 1.013 	imes 10^5) 	imes (2 	imes 10^{-3})}{1400 	imes 8.4}$$T = \frac{1.418 	imes 10^{14} 	imes 2 	imes 10^{-3}}{11760} = \frac{2.836 	imes 10^{11}}{11760} \approx 2.41 	imes 10^7 \, K$Thus, the temperature is approximately $2.4 	imes 10^7 \, K$.

If the density of the Sun is $1.4 \, g \, cm^{-3}$, the pressure is $1.4 \times 10^9 \, \text{atm}$, and the molar mass of the gas in the Sun is $2 \, g \, mol^{-1}$, then the temperature of the Sun will be: [Take $R = 8.4 \, J \, mol^{-1} \, K^{-1}$]

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