Butane (C4 H10) has a heat of vaporization of 22.44 kJ / mol and a normal boiling point of -0.4

Butane (C4 H10) has a heat of vaporization of 22.44 kJ /...

Key Concepts

Phase Equilibrium

In a closed system, phase equilibrium occurs when the rate of evaporation equals the rate of condensation, meaning the vapor and liquid phases coexist stably. This balance is essential to understand how much of a substance remains in either phase when the system is contained and subjected to temperature changes.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It is a fundamental property that increases with temperature and determines the extent to which a liquid will evaporate under specific conditions.

Heat of Vaporization

The heat of vaporization is the energy required to convert one mole of a liquid into its vapor without a temperature change. This concept is crucial when analyzing phase transitions as it relates to the energy balance during vaporization and condensation processes.

Boiling Point

The boiling point is the temperature at which a liquid's vapor pressure equals the surrounding atmospheric pressure, leading to a rapid phase change from liquid to gas. Understanding the boiling point helps predict the phase behavior of a substance under various thermal conditions.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation provides a quantitative relationship between vapor pressure and temperature, incorporating the heat of vaporization. This relationship is useful for predicting how vapor pressure changes as the temperature varies, which is critical for determining phase distributions in a system.

Ideal Gas Law

The ideal gas law relates pressure, volume, temperature, and the number of moles of a gas. It is often used to estimate the behavior and quantity of the gaseous phase in equilibrium problems where the assumptions about ideal gas behavior are applicable.

Transcript

00:01 In this question, we want to determine the mass of liquid putain at negative 2 to 0 degrees celsius.

00:07 So for us to determine the mass, we know that the number of moles of a substance are equal to the mass divided by the molar mass of a substance.

00:14 So since we are dealing with putane, the molar mass is going to be equal to 58.

00:20 Then the whole purpose here will be determining the number of moles that we can use to determine our mass.

00:26 That is the mass is going to be equal to the number of moles multiplied by 58 which is the molar mass and coming to determining the number of moles to simplify this we can assume ideal class behavior where pv is equal to n r t so our n is going to be equal to pv divided by rt so the main goal here is going to be determining our number of moles but we first of all have to determine our pressure because our temperature we already have and our arrow we already have and the volume we're going to be using 2 -50 mils so to determine the pressure to determine the pressure we've got two states state one moving to state number two and these have their own pressures the initial pressure here is one atmosphere we need the pressure at state number two and at state number two we are in at 251 kelvin and we need to know the pressure that we are then going to use to determine the number of modes.

01:35 Now what we can use here to determine the pressure, we know that limb p2 over p1 is equal to the enthalpy change of vaporization divided by our r.

01:45 This is going to be negative, multiply by 1 over t2 minus 1 over t1.

01:51 So we have all this information except for the final pressure, p2 now moving on we can just plug in the values in here to say our link p2 over p1 this is going to be equal to negative to 2 .4 kilojoules per milchols per mole divided by 0 .0 .08314 and this is in kilojoules per mole kelvin multiplied by 1 over 251 minus 1 over 272 .6 so all this link p2 over p1 this is going to be equal to negative 0 .853 so for us to determine to remove this logarithm we raise this as an exponent to say p1 over p2 rather this is p2 over p1 this is equal to the exponent of negative 0 .853 so our p2 p2 is going to be equal to the exponent of negative 0 .853 multiplied by p1 which is one atmosphere so our p2 here is going to be equal to 0 .4 -266 atmosphere this is the vapor pressure 8 negative to 2 to decrease celsius now that we have this information we can then use this to plug into our ideal gas law to determine the number of modes so plugging this information we're going to have the number of moles being equal to number of moles being equal to the pressure 0 .4 to 66 multiplied by the volume which is 0 .25 liters and this is determined from 250 milliliters divided by our r our our r now we have to use 0 .028 2006 because our pressure is in atmosphere so so the r that we're going to use is atmosphere liters per mole kelvin.

04:05 So since we've used this pressure in atmosphere, this is going to be our r multiplied by the temperature which has to be in kelvin and this is 251 in kelvin.

04:16 Therefore the number of moles, the number of moles this is going to be equal to 0 .0518.

04:29 Now, now that we have the number of moles, we can determine the mass, say the mass is equal to number of moles which is 0 .0518 multiplied by 58.

04:42 So the mass, this is equal to the mass equals to 0 .3 grams.

04:55 So what we have here is the mass of liquid, this is equal to the total mass minus the mass of the of cassius butane.

05:05 So the mass of liquid putane this is equal to 0 .55 minus 0 .3.

05:14 So this is going to be equal to the mass of liquid.

05:20 This is going to be equal to 0 .25 grams.

05:24 Right...

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