
A gas occupies a volume of \[300cc\] at \[27^o\]C and \[620mm\] pressure. The volume of gas at \47^o\]C and \[640mm\] pressure is:
A.260 cc
B.310 cc
C.390 cc
D.450 cc
Answer
585.3k+ views
Hint:To answer this question, you should recall the concept of transfer of gases. Use the relation of an ideal gas for the two conditions. Use the given values to find the answer to this question.
Formula used:
Ideal gas equation \[{\text{PV = nRT}}\] where ${\text{P}}$ is pressure, ${\text{V}}$ is volume, ${\text{n}}$ is the number of moles, ${\text{R}}$ is the universal gas constant, ${\text{T}}$ is temperature.
Complete Step by step solution:
We know that as the gas is transferred without the change in the number of moles, they will remain constant in both the conditions. The ideal gas equation for both the states can be written as:
\[\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}\].
We know the values of pressure and temperature which need to be converted into appropriate units.
Substituting these values into this relation we get:
\[\dfrac{{{{\text{V}}_{\text{1}}} \times 640}}{{(273 + 47)}} = \dfrac{{620 \times 300}}{{(273 + 27)}}\]
Rearranging and solving we get the value
\[{{\text{V}}_{{\text{1}}}} = \dfrac{{620 \times 300 \times 320}}{{640 \times 300}}{\text{ }} = 310{\text{cc}}\] .
This is the volume of gas in new conditions.
Therefore, we can conclude that the correct answer to this question is option B.
Note:At 'higher temperature' and 'lower pressure', a gas behaves like an ideal gas, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the space between them. You should remember the applicability of gas laws This law does not apply to real gases. They behave ideally when the gases are at low pressure and high temperature. Hence, at high pressures and low temperatures, gas laws are not applicable since the gases are more likely to react and change the conditions of the system.
Formula used:
Ideal gas equation \[{\text{PV = nRT}}\] where ${\text{P}}$ is pressure, ${\text{V}}$ is volume, ${\text{n}}$ is the number of moles, ${\text{R}}$ is the universal gas constant, ${\text{T}}$ is temperature.
Complete Step by step solution:
We know that as the gas is transferred without the change in the number of moles, they will remain constant in both the conditions. The ideal gas equation for both the states can be written as:
\[\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}\].
We know the values of pressure and temperature which need to be converted into appropriate units.
Substituting these values into this relation we get:
\[\dfrac{{{{\text{V}}_{\text{1}}} \times 640}}{{(273 + 47)}} = \dfrac{{620 \times 300}}{{(273 + 27)}}\]
Rearranging and solving we get the value
\[{{\text{V}}_{{\text{1}}}} = \dfrac{{620 \times 300 \times 320}}{{640 \times 300}}{\text{ }} = 310{\text{cc}}\] .
This is the volume of gas in new conditions.
Therefore, we can conclude that the correct answer to this question is option B.
Note:At 'higher temperature' and 'lower pressure', a gas behaves like an ideal gas, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the space between them. You should remember the applicability of gas laws This law does not apply to real gases. They behave ideally when the gases are at low pressure and high temperature. Hence, at high pressures and low temperatures, gas laws are not applicable since the gases are more likely to react and change the conditions of the system.
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