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Assertion (A) Gases do not liquefy above...

Assertion (A) Gases do not liquefy above their critical temperature, even on applying high pressure.
Reason (R) Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.

A

Both A and R are true and R is the correct explanation of A.

B

Both A and R are true and R is not the correct explanation of A.

C

A is true but R is false.

D

A is false and R is true.

Text Solution

Verified by Experts

The correct Answer is:
A
Reason is the correct explanation for assertion
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Sketch shows the plot of Z v/s P for a hypothetical gas for one mole at three distint temperature. Boyle's temperature is the temperature at which gas shows ideal behaviour over a pressure range in the low pressure region.Boyle's temperature (T_b)=a/(Rb) .If a plot is obtained at temperature 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 CO_2 and the temperature well above critical temperature curve is more like H_2 at 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+(Pb)/(RT) ( R=2 cal "mol"^(-1) K^(-1) ) Which of the following is correct :

Sketch shows the plot of Z v/s P for a hypothetical gas for one mole at three distint temperature. Boyle's temperature is the temperature at which gas shows ideal behaviour over a pressure range in the low pressure region.Boyle's temperature (T_b)=a/(Rb) .If a plot is obtained at temperature 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 CO_2 and the temperature well above critical temperature curve is more like H_2 at 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+(Pb)/(RT) ( R=2 cal "mol"^(-1) K^(-1) ) In very high pressure region if Z v/s P is plotted at 1200 K for the above gas then it will have greatest slope.

Sketch shows the plot of Z v/s P for a hypothetical gas for one mole at three distint temperature. Boyle's temperature is the temperature at which gas shows ideal behaviour over a pressure range in the low pressure region.Boyle's temperature (T_b)=a/(Rb) .If a plot is obtained at temperature 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 CO_2 and the temperature well above critical temperature curve is more like H_2 at 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+(Pb)/(RT) ( R=2 cal "mol"^(-1) K^(-1) ) For 500 K plot value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure ) then the value of b/(RT) will be

Sketch shows the plot of Z vs P for 1 mol of a hypothetical gas at three distinct temperature. Boyle’s temperature is the temperature at which a gas shows ideal behaviour over a pressure range in the low pressure region. Boyle’s temperature (T_(b)) = (a)/(Rb) . 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 like CO_(2) and the temperature well above critical temperature curve is more like H_(2) as shown above. At high pressure suppose all the constant temperature curve varies linearly with pressure according to the following equation: Z =1 + (Pb)/(RT) (R = 2 cal mol^(-1) K^(-1)) For 500 K plot the value of Z changes from 2 to 2.2 if pressure is varied from 1000 atm to 1200 atm (high pressure) then the value of (b)/(RT) will be :