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choli [55]
3 years ago
11

How is the combined gas law used to calculate changes in pressure, temperatures, and/or volume for a fixed amount of gas?

Physics
1 answer:
nalin [4]3 years ago
5 0

Answer:

Let's start by considering the ideal gas law:

pV=nRT

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature

This equation can also be rewritten as

\frac{pV}{T}=nR

Now, if we consider a fixed amount of gas, this means that the number of moles (n) is constant. So we can rewrite the equation as

\frac{pV}{T}=const.

And therefore, if we consider a gas undergoing a certain transformation from 1 to 2, we can write

\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}

where 1 indicates the conditions of the gas at the beginning and 2 the conditions of the gas after the process. So, the change in pressure/temperature/volume of the gas can be found by using this equation.

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(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur
GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, \rho _g = 2.27 kg/m³

density of liquid, \rho _l = 972 kg/m³

speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

P_o is the pressure amplitude

P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

3 0
3 years ago
Ina shoots a large marble (Marble A, mass: 0.08 kg) at a smaller marble (Marble B, mass: 0.05 kg) that is sitting still. Marble
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Answer:

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Explanation:

Momentum is conserved.

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0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v

0.12 kg m/s = (0.05 kg) v

v = 2.4 m/s

4 0
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