A ridged steel tank filled with 62.7L of nitrogen gas of 85.0L atm

Chemistry

1 answer:

mylen [45]2 years ago

8 0

The final gas pressure : 175.53 atm

<h3>Further explanation</h3>

Maybe the complete question is like this :

A ridged steel tank filled with 62.7 l of nitrogen gas at 85.0 atm and 19 °C is heated to 330 °C while the volume remains constant. what is the final gas pressure?

The volume remains constant⇒Gay Lussac's Law

<em>When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature </em>

$\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

P₁=85 atm

T₁=19+273=292 K

T₂=330+273=603 K

$\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{85\times 603}{292}\\\\P_2=175.53~atm$

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Answer:

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A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

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