1
Question
Isotherms of carbon dioxide at various temperatures are represented in Fig. 5.5. Answer the following questions based on this figure.
A) In which state will CO2 exist between the points a and b at temperature T1 ?
B) At what point will CO2 start liquefying when temperature is T1 ?
C) At what point will CO2 be completely liquefied when temperature is T2 ?
D) Will condensation take place when the temperature is T3?
E) What portion of the isotherm at T1 represents liquid and gaseous CO2 at equilibrium?
A) In which state will CO2 exist between the points a and b at temperature T1 ?
B) At what point will CO2 start liquefying when temperature is T1 ?
C) At what point will CO2 be completely liquefied when temperature is T2 ?
D) Will condensation take place when the temperature is T3?
E) What portion of the isotherm at T1 represents liquid and gaseous CO2 at equilibrium?
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Solution
A)
CO2 exists in the gaseous state between the points ‘a′ and ‘b′ at temperature T1.
B)
At temperature T1,CO2 starts liquefying at point ‘b′ as shown in the above graph.
C)
At the temperature T2,CO2 will be completely liquefied at the point ‘g′. Because in the curve at the temperature T2, point ‘f′ to ‘g′ represents the phase where the gas is being converted to liquid and at point ‘g′ all the gas has been converted to liquid.
D)
As stated in the graph:
T3>TC>T2>T1 Temperature T3>TC (critical temperature), so condensation will not take place when the temperature is T3. Because critical temperature is the temperature above which a gas cannot be liquified and T3 is greater than TC.
E)
At the temperature T1 curve, the equilibrium of liquid and gaseous state of CO2 is represented between the point’s ‘b′ and ‘c′. Because between the points ‘b′ and ‘c′, the pressure being the constant and volume of gas decreases till point ‘c′ so between these points CO2 gas partially exists as in liquid and in the gaseous state i.e. existing in equilibrium.
CO2 exists in the gaseous state between the points ‘a′ and ‘b′ at temperature T1.
B)
At temperature T1,CO2 starts liquefying at point ‘b′ as shown in the above graph.
C)
At the temperature T2,CO2 will be completely liquefied at the point ‘g′. Because in the curve at the temperature T2, point ‘f′ to ‘g′ represents the phase where the gas is being converted to liquid and at point ‘g′ all the gas has been converted to liquid.
D)
As stated in the graph:
T3>TC>T2>T1 Temperature T3>TC (critical temperature), so condensation will not take place when the temperature is T3. Because critical temperature is the temperature above which a gas cannot be liquified and T3 is greater than TC.
E)
At the temperature T1 curve, the equilibrium of liquid and gaseous state of CO2 is represented between the point’s ‘b′ and ‘c′. Because between the points ‘b′ and ‘c′, the pressure being the constant and volume of gas decreases till point ‘c′ so between these points CO2 gas partially exists as in liquid and in the gaseous state i.e. existing in equilibrium.
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