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2 years ago

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A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

moved from the tank. What is the pressure of the remaining gas in the tank?

1 answer:

fomenos2 years ago

4 0

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

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Answer:

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now wavelength is given by

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here we know

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Frequency = 493.88 Hz

now wavelength is given by

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here we know

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here we know

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now frequency is given by

$f = \frac{343m/s}{0.5840m}$

f = 587.3 Hz

8. Frequency = speed / wavelength

here we know

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now frequency is given by

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here we know

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here we know

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now wavelength is given by

$\lambda = \frac{343m/s}{659.26}$

$\lambda = 0.520 m$

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