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Compressibility Factor

The curve of ...

Question

The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is /are incorrect (in the low pressure region):

H_{2}

and

H e

show positive deviation from ideal gas equation.

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C O_{2}, C H_{4}

and

O_{2}

show negative deviation from ideal gas equation.

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H_{2}

and

H e

show negative deviation while

C O_{2}, C H_{4}

and

O_{2}

show positive deviation.

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H_{2}

and He are less compressible than that of an ideal gas while

C O_{2}, C H_{4}

and

O_{2}

more compressible than that of ideal gas.

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Solution

The correct option is C $H_{2}$ and $H e$ show negative deviation while $C O_{2}, C H_{4}$ and $O_{2}$ show positive deviation.
We know that, if Z > 1 positive deviation, Z < 1 negative deviation.
From given graph, $H_{2}, H e$ shows positive deviation from ideal gas equation. $C O_{2}, C H_{4}, O_{2}$ shows negative from ideal gas equation.

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Similar questions

Q. The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is /are incorrect (in the low pressure region):

Q. The curve of pressure-volume

(P V)

against pressure

(P)

of the gas at a particular temperature is as shown, according to the graph which of the following is incorrect (in the low-pressure region)?

Q. The sketch shows the plot of Z v/s P for a hypothetical gas for one mole at three distinct temperature . Boyle's temperature is the temperature at which a gas shows ideal behavior over a pressure range in the low-pressure region .Boyle's temperature

(T_{b})

\frac{a}{R b}

.If a plot is obtained at temperatures well below Boyle's temperature then the curve will show negative deviation,in low-pressure region and positive deviation in the high-pressure region.Near critical temperature, the curve is more likely as

C O_{2}

and the temperature well above critical temperature curve is more like

H_{2}

O^{0}

C as shown above.At high pressure suppose all the constant temperature curve varies linearly with pressure according to the following equation

Z = 1 + \frac{P b}{R T} (R = 2 c a l m o l^{- 1} K^{- 1}

)

Which of the following is correct?

The sketch shows the plot of

Z

v/s

P

for a hypothetical gas for one mole at three distinct temperature. Boyle's temperature is the temperature at which a gas shows ideal behavior over a pressure range in the low-pressure region. Boyle's temperature

(T_{b})

\frac{a}{R b}

.If a plot is obtained at temperatures well below Boyle's temperature then the curve will show negative deviation, in the low-pressure region and positive deviation in the high-pressure region. Near critical temperature, the curve is more likely as

C O_{2}

and the temperature well above critical temperature curve is more like

H_{2}

O^{o} C

as shown above.At high pressure suppose all the constant temperature curve varies linearly with pressure according to the following equation

Z = 1 + \frac{P b}{R T} (R = 2 c a l m o l^{- 1} K^{- 1}

)

For 500

K

plot value of

Z

changes from 2 to 2.2 ,if pressure is varied from 1000

a t m

to 1200

a t m

(high pressure) then the value of

\frac{b}{R T}

will be:

Q. An ideal gas in a thermally insulated container, at internal pressure

P_{1},

volume

V_{1}

and absolute temperature

T_{1}

expands irreversibly against zero external pressure so that the pressure of gas becomes

P_{2},

volume

V_{2}

and temperature

T_{2}

. Which of the following statement(s) is / are incorrect?

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Compressibility Factor

Standard XIII Chemistry

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The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is /are incorrect (in the low pressure region):

The correct option is C H2 and He show negative deviation while CO2,CH4 and O2 show positive deviation.We know that, if Z > 1 positive deviation, Z < 1 negative deviation. From given graph, H2, He shows positive deviation from ideal gas equation.CO2, CH4, O2 shows negative from ideal gas equation.