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arsen [322]
2 years ago
5

a gas is placed inside a very cold box . The gas molecules transfer heat to the walls of the box. Which best describes what happ

ens to the gas as it cools down?
Physics
1 answer:
insens350 [35]2 years ago
8 0
As the gas cools down, particle movement slows down with it, and so does the temperature of the gas. This is because the more heat or thermal energy a gas has, the faster the particles move. This is why absolute zero, or 0 degrees Kelvin is defined as zero particle movement, because it has zero energy.
Hope this helps
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The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium
Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz

Given:

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The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz


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Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s
Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) \frac{E_{d} }{m} = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature (T_{1})of 800 ºR and pressure (P_{1}) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{1}) = 191.81 BTu/lb

Specific entropy (s_{1}) = 0.6956 Btu/(lb.ºR)

At the temperature (T_{2})of 600 ºR and pressure (P_{2}) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{2}) = 143.47 BTu/lb

Specific entropy (s_{2}) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}

Q/m = heat transfer = -2 Btu/lb

V_{1} = 400 ft/s

V_{2} = 100 ft/s

\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}[/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]

The equation above is further simplified to:

\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]

\frac{E_{d} }{m} = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0
3 years ago
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