Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.
Clarification:
This process involves transforming gaseous nitrogen into its liquid form.
The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.
Step 1: Calculate the mass of nitrogen gas
Let’s determine the mass of nitrogen gas associated with 297 L.
The density formula is:
With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:
Using these values yields:
The mass of nitrogen gas calculates to be 1356 g.
Step 2: Derive the volume of liquid nitrogen from the mass obtained
The mass for liquid nitrogen remains the same.
With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.
Therefore, the volume of liquid nitrogen is 1.68 L.
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According to the Law, the variation in internal energy (U) of the system is equal to the total of the heat added to the system (q) plus the work performed ON the system (W)
<span>ΔU = q + W </span>
<span>In response to the first question, 0.653 kJ of heat energy is extracted from the system (balloon) while 386 J of work is applied to the balloon, leading to </span>
<span>ΔU = -653J + 386J </span>
<span>=-267J </span>
<span>Thus, the internal energy reduces by 267 J </span>
<span>For the second question, 322 J of heat is supplied to the system (gold bar) while no work is undertaken on the gold bar, marking this as an isochoric/isovolumetric process, thus </span>
<span>ΔU = 322J + 0 </span>
<span>=322J </span>
<span>Hence, internal energy rises by 322 J</span>
The problem provides a conversion factor---> 1 mm3= 7.0 x 10^6 RBC. Therefore, to determine the quantity of red blood cells in your sample, we must first convert Liters to cm3 using the conversion factor--> 1 mL= 1 cm3
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Answer:
Complete Question:
Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.
In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is
ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.
Explanation:
To clarify the answer provided, let’s begin by defining some concepts.
The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.
The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.
The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.
Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.
