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Standard XII

Chemistry

Molar and Specific Heat Capacity

The essential...

Question

The essential conditions for liquefaction of gases were discovered by Andrews in $1869$ as a result of his study of pressure-volume-temperature relationship for $C O_{2}$ . It was found that above a certain temperature, it was impossible to liquefy a gas whatever the pressure was applied. The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature $(T_{c})$ . The pressure required to liquefy a gas at this temperature is called the critical pressure $(P_{c)}$ . The volume occupied by one mole of the substance at the critical temperature and pressure is called critical volume. Critical constants are related with van der Waals' constant as follows:
$V_{c} = 3 b, P_{c} = \frac{a}{27 b^{2}}, T_{c} = \frac{8 a}{27 R b}$ .

The relationship between $P_{C}, V_{C}$ and $T_{C}$ is

P_{C} V_{C} = R T_{C}

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P_{C} V_{C} = 3 R T_{C}

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P_{C} V_{C} = \frac{3}{5} R T_{C}

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P_{C} V_{C} = \frac{3}{8} R T_{C}

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Solution

The correct option is D $P_{C} V_{C} = \frac{3}{8} R T_{C}$
The option (D) is correct answer.
The relationship between $P_{C}$ , $V_{C}$ and $T_{C}$ is $P_{C} V_{C} = \frac{3}{8} R T_{C}$ or $\frac{P_{C} V_{C}}{R T_{C}} = \frac{3}{8}$ or $\frac{R T_{C}}{P_{C} V_{C}} = \frac{8}{3}$

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Similar questions

The essential conditions for liquefaction of gases were discovered by Andrews in

1869

as a result of his study of pressure-volume-temperature relationship for

C O_{2}

. It was found that above a certain temperature, it was impossible to liquefy a gas whatever the pressure was applied. The temperature below which the gas can be liquefied by the application of pressure alone is called critical temperature

(T_{c})

. The pressure required to liquefy a gas at this temperature is called the critical pressure

(P_{c)}

. The volume occupied by one mole of the substance at the critical temperature and pressure is called critical volume. Critical constants are related with van der Waals' constant as follows:

V_{c} = 3 b, P_{c} = \frac{a}{27 b^{2}}, T_{c} = \frac{8 a}{27 R b}

Which of the following parameters is three times the van der Waals' constant

^{'} b^{'}

The critical temperature $(T_{c})$ and critical pressure $p_{c})$ of $C O_{2}$ are ${30.98}^{\circ} C$ and 73 atm respectively.Can $C O_{2}$ (g) be liquefied at $32^{\circ} C$ and 80 atm pressure?

Q. The pressure required to liquefy a gas at the critical temperature is called

Q. If

\frac{P}{P_{c}} = P_{r}, \frac{T}{T_{c}} = T_{r}, a n d \frac{V_{m}}{V_{m, c}} = V_{r}

where

P_{r}

is reduced pressure,

P_{c}

is critical pressure

T_{r}

is reduced temperature,

T_{c}

is critical temperature

V_{r}

is reduced volume,

V_{c}

is critical volume
then the equation of state (or van der Waals equation), only in terms of

P_{r}, T_{r}

and

V_{r}

Consider the following statements.

I. Gaseous substances whose critical temperatures are above room temperatures can be liquefied by applying sufficient pressure to the gas.

II. Gaseous substances whose critical temperatures are below room temperature can be liquefied by the application of pressure and lowering of temperature below $T_{c}$ .

Select the correct alternate.

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The correct option is D PCVC=38RTCThe option (D) is correct answer.The relationship between PC, VC and TC is PCVC=38RTC or PCVCRTC=38 or RTCPCVC=83

The correct option is D $P_{C} V_{C} = \frac{3}{8} R T_{C}$
The option (D) is correct answer.
The relationship between $P_{C}$ , $V_{C}$ and $T_{C}$ is $P_{C} V_{C} = \frac{3}{8} R T_{C}$ or $\frac{P_{C} V_{C}}{R T_{C}} = \frac{3}{8}$ or $\frac{R T_{C}}{P_{C} V_{C}} = \frac{8}{3}$