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7. A cubic box of volume `5.1 times 10^(-2)\ m^3` is filled with air at atmospheric pressure at `20°C`. The box is closed and heated to `180°C`. What is the net force on each side of the box?
7. A cubic box of volume 5.1 times 10^(-2)\ m^3 is filled with air at atmospheric pressure at 20°C. The box is closed and heated to 180°C. What is the net force on each side of the box? `PV = nRT`
`P = F/A`
`F = 7.6times10^3\ N`
Given:
`V = 5.1 times 10^(-2)\ m^3`
`P_1 = 1.00\ atm`
`T_1 = 273 + 20 = 293\ K`
`T_2 = 273 + 180 = 453\ K`
Equation
`PV = nRT`
`P = F/A`
Solution:
Since amount of gas and volume keeps same.
`P/T = (nR)/V = `constant
`=>`
`P_1/T_1 = P_2/T_2`
`P_2 = P_1 T_2/T_1`
`P_2 = (1.00\ atm) (453\ K)/(293\ K)`
`P_2 = 1.55\ atm`
Since it is a cubic box all sides are equal. Area of each side will be
`A = sqrt((5.1 times 10^(-2)\ m^3)^(1/3)) = 0.14\ m^2`
`P = F/A`
`F = PA`
`F = (1.55\ atm)(1.01 times 10^5 Pa)(0.14\ m^2) `
`F = 7.6times10^3\ N` |