The pressure of a gas is 6 times 10^4 N/m^2. | vedclass

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Question

(A) According to the kinetic theory of gases,the pressure $P$ of an ideal gas is related to the kinetic energy per unit volume $(E/V)$ by the formula:

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$9 	imes 10^4 \ J$

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(A) According to the kinetic theory of gases,the pressure $P$ of an ideal gas is related to the kinetic energy per unit volume $(E/V)$ by the formula: $P = \frac{2}{3} \left( \frac{E}{V} ight)$.Rearranging this formula to find the kinetic energy per unit volume,we get: $\frac{E}{V} = \frac{3}{2} P$.Given the pressure $P = 6 	imes 10^4 \ N/m^2$.Substituting the value of $P$ into the equation: $\frac{E}{V} = \frac{3}{2} 	imes (6 	imes 10^4 \ N/m^2)$.$\frac{E}{V} = 3 	imes 3 	imes 10^4 \ J/m^3 = 9 	imes 10^4 \ J/m^3$.

The pressure of a gas is $6 \times 10^4 \ N/m^2$. What will be the kinetic energy per unit volume of the gas molecules?

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