314242
According to kinetic equation, Boyles's law can be expressed as
1 \(\mathrm{\mathrm{PV}=\mathrm{kT}}\)
2 \(\mathrm{\mathrm{PV}=\mathrm{RT}}\)
3 \(\mathrm{P V=\dfrac{3}{2} k T}\)
4 \(\mathrm{P V=\dfrac{2}{3} k T}\)
Explanation:
The product \(\mathrm{P V}\) will have constant value at constant temperature According to Boyle's law.
CHXI06:STATES OF MATTER
314243
The root mean square velocity of \(\mathrm{1 \mathrm{~mol}}\) of a monoatomic gas having molar mass \(\mathrm{\mathrm{M}}\) is \(\mathrm{v_{r m s}}\). The relation between the average kinetic energy (E) of the gas and \(\mathrm{v_{r m s}}\) is
1 \(\mathrm{v_{r m s}=\sqrt{\dfrac{3 E}{2 M}}}\)
2 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{3 M}}}\)
3 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{M}}}\)
4 \(\mathrm{v_{r m s}=\sqrt{\dfrac{E}{3 M}}}\)
Explanation:
\(\mathrm{v_{r m s}=\sqrt{\dfrac{3 R T}{M}}, E=\dfrac{3}{2} R T \Rightarrow R T=\dfrac{2}{3} E}\) \({{\rm{v}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3 \times }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{E}}}}{{\rm{M}}}} {\rm{ = }}\sqrt {\frac{{{\rm{2E}}}}{{\rm{M}}}} \)
CHXI06:STATES OF MATTER
314244
When universal gas constant (R) is divided by Avogadro number N, then the value \(\mathrm{\mathrm{R} / \mathrm{N}}\) is equivalent to
314245
As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
1 Increases
2 Decreases
3 Remains same
4 Becomes half
Explanation:
According to Gay-Lussac's law, temperature of the gas is directly proportional to the pressure of the gas at constant volume.
CHXI06:STATES OF MATTER
314246
Average \(\mathrm{KE}\) of \(\mathrm{CO}_{2}\) at \(27^{\circ} \mathrm{C}\) is E . The average kinetic energy of \(\mathrm{N}_{2}\) at the same temperature will be
1 E
2 22 E
3 \({\rm{E/22}}\)
4 \(3/\sqrt 2 \)
Explanation:
Average KE of a gas depends only on temperature. \({\rm{KE}} = \frac{3}{2}{\rm{kT}}\) \(\therefore \) At the same temperature, \({{\rm{KE}}\,\,{\rm{of C}}{{\rm{O}}_2} = {\rm{KE}}\,{\rm{of }}\,{{\rm{N}}_2} = {\rm{E}}}\)
314242
According to kinetic equation, Boyles's law can be expressed as
1 \(\mathrm{\mathrm{PV}=\mathrm{kT}}\)
2 \(\mathrm{\mathrm{PV}=\mathrm{RT}}\)
3 \(\mathrm{P V=\dfrac{3}{2} k T}\)
4 \(\mathrm{P V=\dfrac{2}{3} k T}\)
Explanation:
The product \(\mathrm{P V}\) will have constant value at constant temperature According to Boyle's law.
CHXI06:STATES OF MATTER
314243
The root mean square velocity of \(\mathrm{1 \mathrm{~mol}}\) of a monoatomic gas having molar mass \(\mathrm{\mathrm{M}}\) is \(\mathrm{v_{r m s}}\). The relation between the average kinetic energy (E) of the gas and \(\mathrm{v_{r m s}}\) is
1 \(\mathrm{v_{r m s}=\sqrt{\dfrac{3 E}{2 M}}}\)
2 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{3 M}}}\)
3 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{M}}}\)
4 \(\mathrm{v_{r m s}=\sqrt{\dfrac{E}{3 M}}}\)
Explanation:
\(\mathrm{v_{r m s}=\sqrt{\dfrac{3 R T}{M}}, E=\dfrac{3}{2} R T \Rightarrow R T=\dfrac{2}{3} E}\) \({{\rm{v}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3 \times }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{E}}}}{{\rm{M}}}} {\rm{ = }}\sqrt {\frac{{{\rm{2E}}}}{{\rm{M}}}} \)
CHXI06:STATES OF MATTER
314244
When universal gas constant (R) is divided by Avogadro number N, then the value \(\mathrm{\mathrm{R} / \mathrm{N}}\) is equivalent to
314245
As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
1 Increases
2 Decreases
3 Remains same
4 Becomes half
Explanation:
According to Gay-Lussac's law, temperature of the gas is directly proportional to the pressure of the gas at constant volume.
CHXI06:STATES OF MATTER
314246
Average \(\mathrm{KE}\) of \(\mathrm{CO}_{2}\) at \(27^{\circ} \mathrm{C}\) is E . The average kinetic energy of \(\mathrm{N}_{2}\) at the same temperature will be
1 E
2 22 E
3 \({\rm{E/22}}\)
4 \(3/\sqrt 2 \)
Explanation:
Average KE of a gas depends only on temperature. \({\rm{KE}} = \frac{3}{2}{\rm{kT}}\) \(\therefore \) At the same temperature, \({{\rm{KE}}\,\,{\rm{of C}}{{\rm{O}}_2} = {\rm{KE}}\,{\rm{of }}\,{{\rm{N}}_2} = {\rm{E}}}\)
314242
According to kinetic equation, Boyles's law can be expressed as
1 \(\mathrm{\mathrm{PV}=\mathrm{kT}}\)
2 \(\mathrm{\mathrm{PV}=\mathrm{RT}}\)
3 \(\mathrm{P V=\dfrac{3}{2} k T}\)
4 \(\mathrm{P V=\dfrac{2}{3} k T}\)
Explanation:
The product \(\mathrm{P V}\) will have constant value at constant temperature According to Boyle's law.
CHXI06:STATES OF MATTER
314243
The root mean square velocity of \(\mathrm{1 \mathrm{~mol}}\) of a monoatomic gas having molar mass \(\mathrm{\mathrm{M}}\) is \(\mathrm{v_{r m s}}\). The relation between the average kinetic energy (E) of the gas and \(\mathrm{v_{r m s}}\) is
1 \(\mathrm{v_{r m s}=\sqrt{\dfrac{3 E}{2 M}}}\)
2 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{3 M}}}\)
3 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{M}}}\)
4 \(\mathrm{v_{r m s}=\sqrt{\dfrac{E}{3 M}}}\)
Explanation:
\(\mathrm{v_{r m s}=\sqrt{\dfrac{3 R T}{M}}, E=\dfrac{3}{2} R T \Rightarrow R T=\dfrac{2}{3} E}\) \({{\rm{v}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3 \times }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{E}}}}{{\rm{M}}}} {\rm{ = }}\sqrt {\frac{{{\rm{2E}}}}{{\rm{M}}}} \)
CHXI06:STATES OF MATTER
314244
When universal gas constant (R) is divided by Avogadro number N, then the value \(\mathrm{\mathrm{R} / \mathrm{N}}\) is equivalent to
314245
As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
1 Increases
2 Decreases
3 Remains same
4 Becomes half
Explanation:
According to Gay-Lussac's law, temperature of the gas is directly proportional to the pressure of the gas at constant volume.
CHXI06:STATES OF MATTER
314246
Average \(\mathrm{KE}\) of \(\mathrm{CO}_{2}\) at \(27^{\circ} \mathrm{C}\) is E . The average kinetic energy of \(\mathrm{N}_{2}\) at the same temperature will be
1 E
2 22 E
3 \({\rm{E/22}}\)
4 \(3/\sqrt 2 \)
Explanation:
Average KE of a gas depends only on temperature. \({\rm{KE}} = \frac{3}{2}{\rm{kT}}\) \(\therefore \) At the same temperature, \({{\rm{KE}}\,\,{\rm{of C}}{{\rm{O}}_2} = {\rm{KE}}\,{\rm{of }}\,{{\rm{N}}_2} = {\rm{E}}}\)
314242
According to kinetic equation, Boyles's law can be expressed as
1 \(\mathrm{\mathrm{PV}=\mathrm{kT}}\)
2 \(\mathrm{\mathrm{PV}=\mathrm{RT}}\)
3 \(\mathrm{P V=\dfrac{3}{2} k T}\)
4 \(\mathrm{P V=\dfrac{2}{3} k T}\)
Explanation:
The product \(\mathrm{P V}\) will have constant value at constant temperature According to Boyle's law.
CHXI06:STATES OF MATTER
314243
The root mean square velocity of \(\mathrm{1 \mathrm{~mol}}\) of a monoatomic gas having molar mass \(\mathrm{\mathrm{M}}\) is \(\mathrm{v_{r m s}}\). The relation between the average kinetic energy (E) of the gas and \(\mathrm{v_{r m s}}\) is
1 \(\mathrm{v_{r m s}=\sqrt{\dfrac{3 E}{2 M}}}\)
2 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{3 M}}}\)
3 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{M}}}\)
4 \(\mathrm{v_{r m s}=\sqrt{\dfrac{E}{3 M}}}\)
Explanation:
\(\mathrm{v_{r m s}=\sqrt{\dfrac{3 R T}{M}}, E=\dfrac{3}{2} R T \Rightarrow R T=\dfrac{2}{3} E}\) \({{\rm{v}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3 \times }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{E}}}}{{\rm{M}}}} {\rm{ = }}\sqrt {\frac{{{\rm{2E}}}}{{\rm{M}}}} \)
CHXI06:STATES OF MATTER
314244
When universal gas constant (R) is divided by Avogadro number N, then the value \(\mathrm{\mathrm{R} / \mathrm{N}}\) is equivalent to
314245
As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
1 Increases
2 Decreases
3 Remains same
4 Becomes half
Explanation:
According to Gay-Lussac's law, temperature of the gas is directly proportional to the pressure of the gas at constant volume.
CHXI06:STATES OF MATTER
314246
Average \(\mathrm{KE}\) of \(\mathrm{CO}_{2}\) at \(27^{\circ} \mathrm{C}\) is E . The average kinetic energy of \(\mathrm{N}_{2}\) at the same temperature will be
1 E
2 22 E
3 \({\rm{E/22}}\)
4 \(3/\sqrt 2 \)
Explanation:
Average KE of a gas depends only on temperature. \({\rm{KE}} = \frac{3}{2}{\rm{kT}}\) \(\therefore \) At the same temperature, \({{\rm{KE}}\,\,{\rm{of C}}{{\rm{O}}_2} = {\rm{KE}}\,{\rm{of }}\,{{\rm{N}}_2} = {\rm{E}}}\)
314242
According to kinetic equation, Boyles's law can be expressed as
1 \(\mathrm{\mathrm{PV}=\mathrm{kT}}\)
2 \(\mathrm{\mathrm{PV}=\mathrm{RT}}\)
3 \(\mathrm{P V=\dfrac{3}{2} k T}\)
4 \(\mathrm{P V=\dfrac{2}{3} k T}\)
Explanation:
The product \(\mathrm{P V}\) will have constant value at constant temperature According to Boyle's law.
CHXI06:STATES OF MATTER
314243
The root mean square velocity of \(\mathrm{1 \mathrm{~mol}}\) of a monoatomic gas having molar mass \(\mathrm{\mathrm{M}}\) is \(\mathrm{v_{r m s}}\). The relation between the average kinetic energy (E) of the gas and \(\mathrm{v_{r m s}}\) is
1 \(\mathrm{v_{r m s}=\sqrt{\dfrac{3 E}{2 M}}}\)
2 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{3 M}}}\)
3 \(\mathrm{v_{r m s}=\sqrt{\dfrac{2 E}{M}}}\)
4 \(\mathrm{v_{r m s}=\sqrt{\dfrac{E}{3 M}}}\)
Explanation:
\(\mathrm{v_{r m s}=\sqrt{\dfrac{3 R T}{M}}, E=\dfrac{3}{2} R T \Rightarrow R T=\dfrac{2}{3} E}\) \({{\rm{v}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3 \times }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{E}}}}{{\rm{M}}}} {\rm{ = }}\sqrt {\frac{{{\rm{2E}}}}{{\rm{M}}}} \)
CHXI06:STATES OF MATTER
314244
When universal gas constant (R) is divided by Avogadro number N, then the value \(\mathrm{\mathrm{R} / \mathrm{N}}\) is equivalent to
314245
As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
1 Increases
2 Decreases
3 Remains same
4 Becomes half
Explanation:
According to Gay-Lussac's law, temperature of the gas is directly proportional to the pressure of the gas at constant volume.
CHXI06:STATES OF MATTER
314246
Average \(\mathrm{KE}\) of \(\mathrm{CO}_{2}\) at \(27^{\circ} \mathrm{C}\) is E . The average kinetic energy of \(\mathrm{N}_{2}\) at the same temperature will be
1 E
2 22 E
3 \({\rm{E/22}}\)
4 \(3/\sqrt 2 \)
Explanation:
Average KE of a gas depends only on temperature. \({\rm{KE}} = \frac{3}{2}{\rm{kT}}\) \(\therefore \) At the same temperature, \({{\rm{KE}}\,\,{\rm{of C}}{{\rm{O}}_2} = {\rm{KE}}\,{\rm{of }}\,{{\rm{N}}_2} = {\rm{E}}}\)
