360290
The ratio of translational and rational kinetic energy at 100 \(K\) temperature is 3:2. Find the internal energy of one mole gas at this temperature
1 \(2075\,J\)
2 \(3120\,J\)
3 \(5128\,J\)
4 \(7126\,J\)
Explanation:
we use \(\frac{{{K_T}}}{{{K_R}}} = \frac{{\frac{3}{2}nRT}}{{\frac{{(f - 3)}}{2}nRT}} = \frac{3}{2} \Rightarrow \frac{3}{{f - 3}} = \frac{3}{2}\) \( \Rightarrow \quad 6 = 3f - 9 \Rightarrow f = 5\) Now \(U=\dfrac{f}{2} n R T=\dfrac{5}{2} \times 1 \times 8.3 \times 100=2075 {~J}\)
PHXI13:KINETIC THEORY
360291
The amount of heat energy required to raise the temperature of \({8 g}\) of Helium at \(NTP\) from \({T_{1} K}\) to \({T_{2} K}\) is
Number of moles in \({\lg {He}=\dfrac{1}{4}}\) Amount of heat energy required to raise its temperature from \({T_{1} K}\) to \({T_{2} K}\) \({=\mu C_{V} \Delta T}\) \({\left(\dfrac{1}{4}\right) \times\left(\dfrac{3}{2} R\right) \times\left(T_{2}-T_{1}\right)}\) \({=\dfrac{3}{8} k_{B} N_{A}\left(T_{2}-T_{1}\right)}\)
PHXI13:KINETIC THEORY
360292
A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from \(300\,\,K\) to \(600\,\,K\), the average K.E. of helium atoms is
1 Halved
2 Increased by factor \(\sqrt{2}\)
3 Doubled
4 Unchanged
Explanation:
Kinetic energy \(\propto\) Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.
PHXI13:KINETIC THEORY
360293
Energy of 10 non rigid diatomic molecules at temperature \({T}\) is
1 \(35\,RT\)
2 \({70 K_{B} T}\)
3 \({\dfrac{7}{2} R T}\)
4 \({35 K_{B} T}\)
Explanation:
For non rigid diatomic molecule, degree of freedom is given by, \({f=7}\) Energy of one molecule \({=\dfrac{f}{2} K_{B} T}\) Energy of 10 molecules \({=10 \times \dfrac{f}{2} K_{B} T=5 \times 7 K_{B} T=35 K_{B} T}\) So, correct option is (4).
360290
The ratio of translational and rational kinetic energy at 100 \(K\) temperature is 3:2. Find the internal energy of one mole gas at this temperature
1 \(2075\,J\)
2 \(3120\,J\)
3 \(5128\,J\)
4 \(7126\,J\)
Explanation:
we use \(\frac{{{K_T}}}{{{K_R}}} = \frac{{\frac{3}{2}nRT}}{{\frac{{(f - 3)}}{2}nRT}} = \frac{3}{2} \Rightarrow \frac{3}{{f - 3}} = \frac{3}{2}\) \( \Rightarrow \quad 6 = 3f - 9 \Rightarrow f = 5\) Now \(U=\dfrac{f}{2} n R T=\dfrac{5}{2} \times 1 \times 8.3 \times 100=2075 {~J}\)
PHXI13:KINETIC THEORY
360291
The amount of heat energy required to raise the temperature of \({8 g}\) of Helium at \(NTP\) from \({T_{1} K}\) to \({T_{2} K}\) is
Number of moles in \({\lg {He}=\dfrac{1}{4}}\) Amount of heat energy required to raise its temperature from \({T_{1} K}\) to \({T_{2} K}\) \({=\mu C_{V} \Delta T}\) \({\left(\dfrac{1}{4}\right) \times\left(\dfrac{3}{2} R\right) \times\left(T_{2}-T_{1}\right)}\) \({=\dfrac{3}{8} k_{B} N_{A}\left(T_{2}-T_{1}\right)}\)
PHXI13:KINETIC THEORY
360292
A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from \(300\,\,K\) to \(600\,\,K\), the average K.E. of helium atoms is
1 Halved
2 Increased by factor \(\sqrt{2}\)
3 Doubled
4 Unchanged
Explanation:
Kinetic energy \(\propto\) Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.
PHXI13:KINETIC THEORY
360293
Energy of 10 non rigid diatomic molecules at temperature \({T}\) is
1 \(35\,RT\)
2 \({70 K_{B} T}\)
3 \({\dfrac{7}{2} R T}\)
4 \({35 K_{B} T}\)
Explanation:
For non rigid diatomic molecule, degree of freedom is given by, \({f=7}\) Energy of one molecule \({=\dfrac{f}{2} K_{B} T}\) Energy of 10 molecules \({=10 \times \dfrac{f}{2} K_{B} T=5 \times 7 K_{B} T=35 K_{B} T}\) So, correct option is (4).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI13:KINETIC THEORY
360290
The ratio of translational and rational kinetic energy at 100 \(K\) temperature is 3:2. Find the internal energy of one mole gas at this temperature
1 \(2075\,J\)
2 \(3120\,J\)
3 \(5128\,J\)
4 \(7126\,J\)
Explanation:
we use \(\frac{{{K_T}}}{{{K_R}}} = \frac{{\frac{3}{2}nRT}}{{\frac{{(f - 3)}}{2}nRT}} = \frac{3}{2} \Rightarrow \frac{3}{{f - 3}} = \frac{3}{2}\) \( \Rightarrow \quad 6 = 3f - 9 \Rightarrow f = 5\) Now \(U=\dfrac{f}{2} n R T=\dfrac{5}{2} \times 1 \times 8.3 \times 100=2075 {~J}\)
PHXI13:KINETIC THEORY
360291
The amount of heat energy required to raise the temperature of \({8 g}\) of Helium at \(NTP\) from \({T_{1} K}\) to \({T_{2} K}\) is
Number of moles in \({\lg {He}=\dfrac{1}{4}}\) Amount of heat energy required to raise its temperature from \({T_{1} K}\) to \({T_{2} K}\) \({=\mu C_{V} \Delta T}\) \({\left(\dfrac{1}{4}\right) \times\left(\dfrac{3}{2} R\right) \times\left(T_{2}-T_{1}\right)}\) \({=\dfrac{3}{8} k_{B} N_{A}\left(T_{2}-T_{1}\right)}\)
PHXI13:KINETIC THEORY
360292
A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from \(300\,\,K\) to \(600\,\,K\), the average K.E. of helium atoms is
1 Halved
2 Increased by factor \(\sqrt{2}\)
3 Doubled
4 Unchanged
Explanation:
Kinetic energy \(\propto\) Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.
PHXI13:KINETIC THEORY
360293
Energy of 10 non rigid diatomic molecules at temperature \({T}\) is
1 \(35\,RT\)
2 \({70 K_{B} T}\)
3 \({\dfrac{7}{2} R T}\)
4 \({35 K_{B} T}\)
Explanation:
For non rigid diatomic molecule, degree of freedom is given by, \({f=7}\) Energy of one molecule \({=\dfrac{f}{2} K_{B} T}\) Energy of 10 molecules \({=10 \times \dfrac{f}{2} K_{B} T=5 \times 7 K_{B} T=35 K_{B} T}\) So, correct option is (4).
360290
The ratio of translational and rational kinetic energy at 100 \(K\) temperature is 3:2. Find the internal energy of one mole gas at this temperature
1 \(2075\,J\)
2 \(3120\,J\)
3 \(5128\,J\)
4 \(7126\,J\)
Explanation:
we use \(\frac{{{K_T}}}{{{K_R}}} = \frac{{\frac{3}{2}nRT}}{{\frac{{(f - 3)}}{2}nRT}} = \frac{3}{2} \Rightarrow \frac{3}{{f - 3}} = \frac{3}{2}\) \( \Rightarrow \quad 6 = 3f - 9 \Rightarrow f = 5\) Now \(U=\dfrac{f}{2} n R T=\dfrac{5}{2} \times 1 \times 8.3 \times 100=2075 {~J}\)
PHXI13:KINETIC THEORY
360291
The amount of heat energy required to raise the temperature of \({8 g}\) of Helium at \(NTP\) from \({T_{1} K}\) to \({T_{2} K}\) is
Number of moles in \({\lg {He}=\dfrac{1}{4}}\) Amount of heat energy required to raise its temperature from \({T_{1} K}\) to \({T_{2} K}\) \({=\mu C_{V} \Delta T}\) \({\left(\dfrac{1}{4}\right) \times\left(\dfrac{3}{2} R\right) \times\left(T_{2}-T_{1}\right)}\) \({=\dfrac{3}{8} k_{B} N_{A}\left(T_{2}-T_{1}\right)}\)
PHXI13:KINETIC THEORY
360292
A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from \(300\,\,K\) to \(600\,\,K\), the average K.E. of helium atoms is
1 Halved
2 Increased by factor \(\sqrt{2}\)
3 Doubled
4 Unchanged
Explanation:
Kinetic energy \(\propto\) Temperature. Hence if temperature is doubles, kinetic energy will also be doubled.
PHXI13:KINETIC THEORY
360293
Energy of 10 non rigid diatomic molecules at temperature \({T}\) is
1 \(35\,RT\)
2 \({70 K_{B} T}\)
3 \({\dfrac{7}{2} R T}\)
4 \({35 K_{B} T}\)
Explanation:
For non rigid diatomic molecule, degree of freedom is given by, \({f=7}\) Energy of one molecule \({=\dfrac{f}{2} K_{B} T}\) Energy of 10 molecules \({=10 \times \dfrac{f}{2} K_{B} T=5 \times 7 K_{B} T=35 K_{B} T}\) So, correct option is (4).