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Taya2010 [7]
3 years ago
5

If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m

agnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120∘C? Assume that the temperature of the air outside the pressure cooker is 20∘C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be pa
Physics
1 answer:
Xelga [282]3 years ago
4 0

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is \frac{135.9}{A}N

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = \frac{Pressure}{Area} we have

Force = F_{120}=\frac{135.9}{A}N.

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Consider two ideal gases, A and B, at the same temperature. The rms speed of the molecules of gas A is twice that ofgas B. How d
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To correctly estimate the average velocity, you must take the squares of the mean velocity and take the square root of this value. This is known as the root mean square (rms) velocity and is shown as follows:

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M – Gas’s molar mass

R – Molar mass constant

T – Temperature (in Kelvin)

Given data is rms speed for gas molecule A is twice that of gas molecule B. So,

                 \left(V_{r m s}\right)_{A}=2\left(V_{r m s}\right)_{B}

Therefore, equating the molecule’s rms speed formula for both A and B,

                  \sqrt{\frac{3 R T}{M_{A}}}=2(\sqrt{\frac{3 R T}{M_{B}}})

On squaring both sides, we get,

                 \frac{3 R T}{M_{A}}=4\left(\frac{3 R T}{M_{B}}\right)

By solving the above equations, we get,

                 M_{A}=\frac{M_{B}}{4}

8 0
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