The temperature of a gas having 2.0 times 10^ | vedclass

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(A) The ideal gas equation in terms of number density $n = N/V$ is given by $P = nkT$.Given:Number density $n = 2.0 	imes 10^{25} 	ext{ m}^{-3}$Pres

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(A) The ideal gas equation in terms of number density $n = N/V$ is given by $P = nkT$.Given:Number density $n = 2.0 	imes 10^{25} 	ext{ m}^{-3}$Pressure $P = 1.38 	ext{ atm} = 1.38 	imes 1.01325 	imes 10^5 	ext{ Pa} \approx 1.4 	imes 10^5 	ext{ Pa}$Boltzmann constant $k = 1.38 	imes 10^{-23} 	ext{ J K}^{-1}$Using the formula $P = nkT$:$T = \frac{P}{nk}$$T = \frac{1.38 	imes 1.01325 	imes 10^5}{2.0 	imes 10^{25} 	imes 1.38 	imes 10^{-23}}$$T = \frac{1.01325 	imes 10^5}{2.0 	imes 10^2}$$T = \frac{101325}{200} \approx 506.6 	ext{ K}$Rounding to the nearest given option,the temperature is approximately $500 	ext{ K}$.

The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 \text{ atm}$ (Given,$k = 1.38 \times 10^{-23} \text{ J K}^{-1}$) is: (in $\text{ K}$)

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