139271
The number of molecules in a gas at pressure $1.64 \times 10^{-3} \mathrm{~atm}$ and temperature $200 \mathrm{~K}$ having the volume $1 \mathrm{cc}$ are
139272
The root mean square and most probable speed of the molecules in a gas are
1 same
2 different
3 cannot say
4 depends on nature of the gas
Explanation:
B Density of two diatomic gases $=\mathrm{d}_{1} \& \mathrm{~d}_{2}$ At same temperature, $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{1}}}}{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{2}}}}$ $=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $=\sqrt{\frac{\mathrm{Vd}_{2}}{\mathrm{Vd}_{1}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\text {rms }}}{\left(\mathrm{v}_{2}\right)_{\text {rms }}}=\sqrt{\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}}$
Hence
Kinetic Theory of Gases
139273
If at the same temperature and pressure, the densities of two diatomic gases are $d_{1}$ and $d_{2}$ respectively, the ratio of mean kinetic energy per molecules of gases will be
1 $1: 1$
2 $\mathrm{d}_{1}: \mathrm{d}_{2}$
3 $\sqrt{\mathrm{d}_{1}}: \sqrt{\mathrm{d}_{2}}$
4 $\sqrt{\mathrm{d}_{2}}: \sqrt{\mathrm{d}_{1}}$
Explanation:
A Pressure and Temperature is same, Densities of two diatomic gases $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ $\because$ mean kinetic energy (K.E.) $=\frac{3}{2} \mathrm{KT}$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{\frac{3}{2} \mathrm{KT}_{1}}{\frac{3}{2} \mathrm{KT}_{2}}$ $\left[\because \mathrm{T}_{1}=\mathrm{T}_{2}\right]$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{1}{1}$
139271
The number of molecules in a gas at pressure $1.64 \times 10^{-3} \mathrm{~atm}$ and temperature $200 \mathrm{~K}$ having the volume $1 \mathrm{cc}$ are
139272
The root mean square and most probable speed of the molecules in a gas are
1 same
2 different
3 cannot say
4 depends on nature of the gas
Explanation:
B Density of two diatomic gases $=\mathrm{d}_{1} \& \mathrm{~d}_{2}$ At same temperature, $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{1}}}}{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{2}}}}$ $=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $=\sqrt{\frac{\mathrm{Vd}_{2}}{\mathrm{Vd}_{1}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\text {rms }}}{\left(\mathrm{v}_{2}\right)_{\text {rms }}}=\sqrt{\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}}$
Hence
Kinetic Theory of Gases
139273
If at the same temperature and pressure, the densities of two diatomic gases are $d_{1}$ and $d_{2}$ respectively, the ratio of mean kinetic energy per molecules of gases will be
1 $1: 1$
2 $\mathrm{d}_{1}: \mathrm{d}_{2}$
3 $\sqrt{\mathrm{d}_{1}}: \sqrt{\mathrm{d}_{2}}$
4 $\sqrt{\mathrm{d}_{2}}: \sqrt{\mathrm{d}_{1}}$
Explanation:
A Pressure and Temperature is same, Densities of two diatomic gases $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ $\because$ mean kinetic energy (K.E.) $=\frac{3}{2} \mathrm{KT}$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{\frac{3}{2} \mathrm{KT}_{1}}{\frac{3}{2} \mathrm{KT}_{2}}$ $\left[\because \mathrm{T}_{1}=\mathrm{T}_{2}\right]$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{1}{1}$
139271
The number of molecules in a gas at pressure $1.64 \times 10^{-3} \mathrm{~atm}$ and temperature $200 \mathrm{~K}$ having the volume $1 \mathrm{cc}$ are
139272
The root mean square and most probable speed of the molecules in a gas are
1 same
2 different
3 cannot say
4 depends on nature of the gas
Explanation:
B Density of two diatomic gases $=\mathrm{d}_{1} \& \mathrm{~d}_{2}$ At same temperature, $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{1}}}}{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{2}}}}$ $=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $=\sqrt{\frac{\mathrm{Vd}_{2}}{\mathrm{Vd}_{1}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\text {rms }}}{\left(\mathrm{v}_{2}\right)_{\text {rms }}}=\sqrt{\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}}$
Hence
Kinetic Theory of Gases
139273
If at the same temperature and pressure, the densities of two diatomic gases are $d_{1}$ and $d_{2}$ respectively, the ratio of mean kinetic energy per molecules of gases will be
1 $1: 1$
2 $\mathrm{d}_{1}: \mathrm{d}_{2}$
3 $\sqrt{\mathrm{d}_{1}}: \sqrt{\mathrm{d}_{2}}$
4 $\sqrt{\mathrm{d}_{2}}: \sqrt{\mathrm{d}_{1}}$
Explanation:
A Pressure and Temperature is same, Densities of two diatomic gases $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ $\because$ mean kinetic energy (K.E.) $=\frac{3}{2} \mathrm{KT}$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{\frac{3}{2} \mathrm{KT}_{1}}{\frac{3}{2} \mathrm{KT}_{2}}$ $\left[\because \mathrm{T}_{1}=\mathrm{T}_{2}\right]$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{1}{1}$
139271
The number of molecules in a gas at pressure $1.64 \times 10^{-3} \mathrm{~atm}$ and temperature $200 \mathrm{~K}$ having the volume $1 \mathrm{cc}$ are
139272
The root mean square and most probable speed of the molecules in a gas are
1 same
2 different
3 cannot say
4 depends on nature of the gas
Explanation:
B Density of two diatomic gases $=\mathrm{d}_{1} \& \mathrm{~d}_{2}$ At same temperature, $\because \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{1}}}}{\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{2}}}}$ $=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $=\sqrt{\frac{\mathrm{Vd}_{2}}{\mathrm{Vd}_{1}}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\text {rms }}}{\left(\mathrm{v}_{2}\right)_{\text {rms }}}=\sqrt{\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}}$
Hence
Kinetic Theory of Gases
139273
If at the same temperature and pressure, the densities of two diatomic gases are $d_{1}$ and $d_{2}$ respectively, the ratio of mean kinetic energy per molecules of gases will be
1 $1: 1$
2 $\mathrm{d}_{1}: \mathrm{d}_{2}$
3 $\sqrt{\mathrm{d}_{1}}: \sqrt{\mathrm{d}_{2}}$
4 $\sqrt{\mathrm{d}_{2}}: \sqrt{\mathrm{d}_{1}}$
Explanation:
A Pressure and Temperature is same, Densities of two diatomic gases $=\mathrm{d}_{1}$ and $\mathrm{d}_{2}$ $\because$ mean kinetic energy (K.E.) $=\frac{3}{2} \mathrm{KT}$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{\frac{3}{2} \mathrm{KT}_{1}}{\frac{3}{2} \mathrm{KT}_{2}}$ $\left[\because \mathrm{T}_{1}=\mathrm{T}_{2}\right]$ $\frac{(\text { K.E. })_{1}}{(\text { K.E. })_{2}}=\frac{1}{1}$
