
When \[0.25L\] of liquid nitrogen \[(d = 0.807gm{l^{ - 1}})\] is vaporized, what volume does the resulting gas occupy at \[{25^ \circ }C\] and \[5.00atm\] ?
Options-
A.\[71L\]
B.\[54L\]
C.\[35L\]
D.\[32L\]
Answer
502.2k+ views
Hint: 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.
\[mass = density \times volume\]
\[mass({N_2}) = 0.807gm{l^{ - 1}} \times 0.25 \times 1000ml = 201.75g\]
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
\[{\text{number of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\]
\[{\text{number of moles}}({N_2}) = \dfrac{{201.75g}}{{28.0gmo{l^{ - 1}}}} = 7.205mol\]
Using the ideal gas law, for calculating the volume of nitrogen gas by inserting the value of temperature and pressure,
\[V = \dfrac{{nRT}}{P}\]
\[V({N_2}) = \dfrac{{7.205mol \times 0.0821Latm{K^{ - 1}}mo{l^{ - 1}} \times 298K}}{{5.00atm}} \approx 35L\]
Hence, the volume of nitrogen gas is \[35L\] 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.
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.
\[mass = density \times volume\]
\[mass({N_2}) = 0.807gm{l^{ - 1}} \times 0.25 \times 1000ml = 201.75g\]
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
\[{\text{number of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\]
\[{\text{number of moles}}({N_2}) = \dfrac{{201.75g}}{{28.0gmo{l^{ - 1}}}} = 7.205mol\]
Using the ideal gas law, for calculating the volume of nitrogen gas by inserting the value of temperature and pressure,
\[V = \dfrac{{nRT}}{P}\]
\[V({N_2}) = \dfrac{{7.205mol \times 0.0821Latm{K^{ - 1}}mo{l^{ - 1}} \times 298K}}{{5.00atm}} \approx 35L\]
Hence, the volume of nitrogen gas is \[35L\] 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.
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