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3 years ago

11

A 1.50 liter flask at a temperature of 25°C contains a mixture of 0.158 moles of methane, 0.09 moles of ethane, and 0.044 moles

of butane. What is the total pressure of the mixture inside the flask?

1 answer:

Julli [10]3 years ago

7 0

Applying the general gas equation : PV=nRT
V = 1.50 litres
R = <span>0.08205746 L atm/K mol
T = </span><span>25°C = 25+273 = 298 K
n = </span>0.158 moles+0.09 moles+<span> 0.044 moles = 0.292 moles
</span><span>Re arranging the equation:
P = nRT / V = (0.158 + 0.09 + 0.044) mol x (0.08205746 L atm/K mol)
x (25 + 273)K / (1.50 L)
= 4.76 atm
= 482 kPa
= 3617 mmHg</span>

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Given:

sound level, $\beta= 101 dB$

Area, A = $22\;m^{2}$

Time, $\triangle t = 20\;min=1200\;s$

Intensity, $I=1\times 10^{-12}\;W/m^{2}$

$r=1\;km=1000\;m$

(a)

We know that, Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Solving the above equation for sound intensity,

$I=I_{o} \times 10^{\frac{\beta}{10} }$

$I=1 \times 10^{-12} \times 10^{\frac{101}{10} }$

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Therefore, The sound energy is,

$E=P\times \triangle t$

Substitute $P=I \times A$ in the above equation,

$E=I \times A \times \triangle t$

$E=0.0126 \times 22 \times 1200$

$E=332.6\;J$

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Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

$A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi$

Substitute the known value in the above equation ,

$A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi$

$A_{hemisphere} = 6283185\;m^{2}$

Sound Intensity is,

$I = \frac{P}{A_{hemisphere}}$

Substitute $P=I \times A$ in the above equation,

$I = \frac{I \times A}{A_{hemisphere}}$

Substitute the known value in the above equation,

$I = \frac{0.0126 \times 22}{6283185}$

$I = 4.4 \times 10^{-8}\;W/m^{2}$

Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Substitute the known value in the above equation,

$\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )$

$\beta=46.4\;dB$

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