Solveeit Logo

Question

Question: When \[0.25L\] of liquid nitrogen \[(d = 0.807gm{l^{ - 1}})\] is vaporized, what volume does the res...

When 0.25L0.25L of liquid nitrogen (d=0.807gml1)(d = 0.807gm{l^{ - 1}}) is vaporized, what volume does the resulting gas occupy at 25C{25^ \circ }C and 5.00atm5.00atm ?
Options-
A.71L71L
B.54L54L
C.35L35L
D.32L32L

Explanation

Solution

The density of any liquid is the ratio of its mass and volume, therefore knowing the density and the volume of any liquid one can easily determine the mass contained in that particular volume. Once the magnitude of mass is known, it can be used to determine the amount of the substance.

Complete answer:
The process of evaporation involves the phase transformation from the liquid state to the gaseous state of a particular substance.
The evaporation of liquid nitrogen gives the gaseous nitrogen. The mass and the amount of nitrogen remains constant throughout the phase transformation (as it is a physical process). The pressure and temperature also remain unchanged and only the volume is increased as gases have a tendency to expand. (The expansion of gases is associated with the weak forces of attraction present in between their particles).
Thus the density and volume of liquid nitrogen can be used to determine its mass.
mass=density×volumemass = density \times volume
mass(N2)=0.807gml1×0.25×1000ml=201.75gmass({N_2}) = 0.807gm{l^{ - 1}} \times 0.25 \times 1000ml = 201.75g
Using the mass of nitrogen present in the given amount of volume, the number of moles can be found out by putting in the value of the molar mass of nitrogen
number of moles=given massmolar mass{\text{number of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}
number of moles(N2)=201.75g28.0gmol1=7.205mol{\text{number of moles}}({N_2}) = \dfrac{{201.75g}}{{28.0gmo{l^{ - 1}}}} = 7.205mol
Using the ideal gas law, for calculating the volume of nitrogen gas by inserting the value of temperature and pressure,
V=nRTPV = \dfrac{{nRT}}{P}
V(N2)=7.205mol×0.0821LatmK1mol1×298K5.00atm35LV({N_2}) = \dfrac{{7.205mol \times 0.0821Latm{K^{ - 1}}mo{l^{ - 1}} \times 298K}}{{5.00atm}} \approx 35L

Hence, the volume of nitrogen gas is 35L35L and the correct option is option (c).

Note:
The nitrogen gas that is formed from the evaporation of its liquid state results in diatomic nitrogen molecules containing strong triple bonds. It is assumed that the nitrogen gas formed behaves ideally at the given conditions of temperature and pressure.