139019
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2: 3$. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature is $4: 3$. The ratio of their densities is
139020
Pressure (P) versus density (d) graph of an ideal gas is shown in the figure. Choose the correct statement. During the process $A B$, workdone by the gas is positive
1 During the process $\mathrm{AB}$, workdone by the gas is positive.
2 During the process $\mathrm{AB}$, workdone by the gas is negative.
3 During the process BC, internal energy of the gas increases.
4 During the process DA, internal energy of the gas remains constant.
Explanation:
D Density (d) $=\frac{\mathrm{m}}{\mathrm{V}}$ So, mass is constant then graph $\Rightarrow\left(\mathrm{P} \propto \frac{1}{\mathrm{~V}}\right)$ $(a, b)$ During $\mathrm{AB}$ process volume constant then, $\mathrm{W}=0$ So, neither positive nor negative in $\mathrm{AB}$ process (c) In BC process $\mathrm{P} \propto \mathrm{d}$ and $\rho \propto \frac{1}{\mathrm{~V}}$ (d) Then $\mathrm{P} \propto \frac{1}{\mathrm{~V}}, \quad \mathrm{PV}=$ Constant Therefore $\mathrm{PV}=\mathrm{nRT}$ So, Temperature is constant in BC process Therefore Internal Energy remains constant
Shift-I]
Kinetic Theory of Gases
139021
A sample of an ideal gas $(\gamma=1.4)$ is heated at constant pressure. If $80 \mathrm{~J}$ of heat is supplied to the gas. the workdone by the gas is
1 $12.80 \mathrm{~J}$
2 $28.62 \mathrm{~J}$
3 $26.28 \mathrm{~J}$
4 $22.86 \mathrm{~J}$
Explanation:
D Given that, $\Delta \mathrm{Q}=80$ Joule At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}$ and $\mathrm{dU}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$ By thermodynamics first law So, $\Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}} \Rightarrow 1-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \quad\left\{\because \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma\right\}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=1-\frac{1}{\gamma}=1-\frac{1}{1.4}=\frac{2}{7}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{2}{7} \Rightarrow \Delta \mathrm{W}=\frac{2}{7} \Delta \mathrm{Q}$ $\Delta \mathrm{W}=\frac{2}{7} \times 80=22.85$ Joule $\Delta \mathrm{W}=22.86$ Joule
Shift-II
Kinetic Theory of Gases
139022
A graph between pressure $P$ (along $y$-axis) and absolute temperature, $T$ (along $x$-axis) for equal moles of two gases has been drawn. Given that volume of second gas is more than volume of first gas. Which of the following statement is correct?
1 Slope of gas 1 is less than gas 2
2 Slope of gas 1 is more than gas 2
3 Both have some slopes
4 None of the above
Explanation:
B According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{V}}$ $\frac{\mathrm{P}}{\mathrm{T}}$ represents slop of the graph as the number of moles are the same for the two gases. Then $\quad \frac{\mathrm{P}}{\mathrm{T}} \propto \frac{1}{\mathrm{~V}}$ Given, $\quad \mathrm{V}_{2}>\mathrm{V}_{1}$ slope $_{1}>$ slope $_{2}$
139019
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2: 3$. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature is $4: 3$. The ratio of their densities is
139020
Pressure (P) versus density (d) graph of an ideal gas is shown in the figure. Choose the correct statement. During the process $A B$, workdone by the gas is positive
1 During the process $\mathrm{AB}$, workdone by the gas is positive.
2 During the process $\mathrm{AB}$, workdone by the gas is negative.
3 During the process BC, internal energy of the gas increases.
4 During the process DA, internal energy of the gas remains constant.
Explanation:
D Density (d) $=\frac{\mathrm{m}}{\mathrm{V}}$ So, mass is constant then graph $\Rightarrow\left(\mathrm{P} \propto \frac{1}{\mathrm{~V}}\right)$ $(a, b)$ During $\mathrm{AB}$ process volume constant then, $\mathrm{W}=0$ So, neither positive nor negative in $\mathrm{AB}$ process (c) In BC process $\mathrm{P} \propto \mathrm{d}$ and $\rho \propto \frac{1}{\mathrm{~V}}$ (d) Then $\mathrm{P} \propto \frac{1}{\mathrm{~V}}, \quad \mathrm{PV}=$ Constant Therefore $\mathrm{PV}=\mathrm{nRT}$ So, Temperature is constant in BC process Therefore Internal Energy remains constant
Shift-I]
Kinetic Theory of Gases
139021
A sample of an ideal gas $(\gamma=1.4)$ is heated at constant pressure. If $80 \mathrm{~J}$ of heat is supplied to the gas. the workdone by the gas is
1 $12.80 \mathrm{~J}$
2 $28.62 \mathrm{~J}$
3 $26.28 \mathrm{~J}$
4 $22.86 \mathrm{~J}$
Explanation:
D Given that, $\Delta \mathrm{Q}=80$ Joule At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}$ and $\mathrm{dU}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$ By thermodynamics first law So, $\Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}} \Rightarrow 1-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \quad\left\{\because \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma\right\}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=1-\frac{1}{\gamma}=1-\frac{1}{1.4}=\frac{2}{7}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{2}{7} \Rightarrow \Delta \mathrm{W}=\frac{2}{7} \Delta \mathrm{Q}$ $\Delta \mathrm{W}=\frac{2}{7} \times 80=22.85$ Joule $\Delta \mathrm{W}=22.86$ Joule
Shift-II
Kinetic Theory of Gases
139022
A graph between pressure $P$ (along $y$-axis) and absolute temperature, $T$ (along $x$-axis) for equal moles of two gases has been drawn. Given that volume of second gas is more than volume of first gas. Which of the following statement is correct?
1 Slope of gas 1 is less than gas 2
2 Slope of gas 1 is more than gas 2
3 Both have some slopes
4 None of the above
Explanation:
B According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{V}}$ $\frac{\mathrm{P}}{\mathrm{T}}$ represents slop of the graph as the number of moles are the same for the two gases. Then $\quad \frac{\mathrm{P}}{\mathrm{T}} \propto \frac{1}{\mathrm{~V}}$ Given, $\quad \mathrm{V}_{2}>\mathrm{V}_{1}$ slope $_{1}>$ slope $_{2}$
139019
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2: 3$. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature is $4: 3$. The ratio of their densities is
139020
Pressure (P) versus density (d) graph of an ideal gas is shown in the figure. Choose the correct statement. During the process $A B$, workdone by the gas is positive
1 During the process $\mathrm{AB}$, workdone by the gas is positive.
2 During the process $\mathrm{AB}$, workdone by the gas is negative.
3 During the process BC, internal energy of the gas increases.
4 During the process DA, internal energy of the gas remains constant.
Explanation:
D Density (d) $=\frac{\mathrm{m}}{\mathrm{V}}$ So, mass is constant then graph $\Rightarrow\left(\mathrm{P} \propto \frac{1}{\mathrm{~V}}\right)$ $(a, b)$ During $\mathrm{AB}$ process volume constant then, $\mathrm{W}=0$ So, neither positive nor negative in $\mathrm{AB}$ process (c) In BC process $\mathrm{P} \propto \mathrm{d}$ and $\rho \propto \frac{1}{\mathrm{~V}}$ (d) Then $\mathrm{P} \propto \frac{1}{\mathrm{~V}}, \quad \mathrm{PV}=$ Constant Therefore $\mathrm{PV}=\mathrm{nRT}$ So, Temperature is constant in BC process Therefore Internal Energy remains constant
Shift-I]
Kinetic Theory of Gases
139021
A sample of an ideal gas $(\gamma=1.4)$ is heated at constant pressure. If $80 \mathrm{~J}$ of heat is supplied to the gas. the workdone by the gas is
1 $12.80 \mathrm{~J}$
2 $28.62 \mathrm{~J}$
3 $26.28 \mathrm{~J}$
4 $22.86 \mathrm{~J}$
Explanation:
D Given that, $\Delta \mathrm{Q}=80$ Joule At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}$ and $\mathrm{dU}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$ By thermodynamics first law So, $\Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}} \Rightarrow 1-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \quad\left\{\because \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma\right\}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=1-\frac{1}{\gamma}=1-\frac{1}{1.4}=\frac{2}{7}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{2}{7} \Rightarrow \Delta \mathrm{W}=\frac{2}{7} \Delta \mathrm{Q}$ $\Delta \mathrm{W}=\frac{2}{7} \times 80=22.85$ Joule $\Delta \mathrm{W}=22.86$ Joule
Shift-II
Kinetic Theory of Gases
139022
A graph between pressure $P$ (along $y$-axis) and absolute temperature, $T$ (along $x$-axis) for equal moles of two gases has been drawn. Given that volume of second gas is more than volume of first gas. Which of the following statement is correct?
1 Slope of gas 1 is less than gas 2
2 Slope of gas 1 is more than gas 2
3 Both have some slopes
4 None of the above
Explanation:
B According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{V}}$ $\frac{\mathrm{P}}{\mathrm{T}}$ represents slop of the graph as the number of moles are the same for the two gases. Then $\quad \frac{\mathrm{P}}{\mathrm{T}} \propto \frac{1}{\mathrm{~V}}$ Given, $\quad \mathrm{V}_{2}>\mathrm{V}_{1}$ slope $_{1}>$ slope $_{2}$
139019
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2: 3$. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature is $4: 3$. The ratio of their densities is
139020
Pressure (P) versus density (d) graph of an ideal gas is shown in the figure. Choose the correct statement. During the process $A B$, workdone by the gas is positive
1 During the process $\mathrm{AB}$, workdone by the gas is positive.
2 During the process $\mathrm{AB}$, workdone by the gas is negative.
3 During the process BC, internal energy of the gas increases.
4 During the process DA, internal energy of the gas remains constant.
Explanation:
D Density (d) $=\frac{\mathrm{m}}{\mathrm{V}}$ So, mass is constant then graph $\Rightarrow\left(\mathrm{P} \propto \frac{1}{\mathrm{~V}}\right)$ $(a, b)$ During $\mathrm{AB}$ process volume constant then, $\mathrm{W}=0$ So, neither positive nor negative in $\mathrm{AB}$ process (c) In BC process $\mathrm{P} \propto \mathrm{d}$ and $\rho \propto \frac{1}{\mathrm{~V}}$ (d) Then $\mathrm{P} \propto \frac{1}{\mathrm{~V}}, \quad \mathrm{PV}=$ Constant Therefore $\mathrm{PV}=\mathrm{nRT}$ So, Temperature is constant in BC process Therefore Internal Energy remains constant
Shift-I]
Kinetic Theory of Gases
139021
A sample of an ideal gas $(\gamma=1.4)$ is heated at constant pressure. If $80 \mathrm{~J}$ of heat is supplied to the gas. the workdone by the gas is
1 $12.80 \mathrm{~J}$
2 $28.62 \mathrm{~J}$
3 $26.28 \mathrm{~J}$
4 $22.86 \mathrm{~J}$
Explanation:
D Given that, $\Delta \mathrm{Q}=80$ Joule At constant pressure $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}$ and $\mathrm{dU}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$ By thermodynamics first law So, $\Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{\mathrm{n}\left(\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}\right) \Delta \mathrm{T}}{\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}} \Rightarrow 1-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \quad\left\{\because \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma\right\}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=1-\frac{1}{\gamma}=1-\frac{1}{1.4}=\frac{2}{7}$ $\frac{\Delta \mathrm{W}}{\Delta \mathrm{Q}}=\frac{2}{7} \Rightarrow \Delta \mathrm{W}=\frac{2}{7} \Delta \mathrm{Q}$ $\Delta \mathrm{W}=\frac{2}{7} \times 80=22.85$ Joule $\Delta \mathrm{W}=22.86$ Joule
Shift-II
Kinetic Theory of Gases
139022
A graph between pressure $P$ (along $y$-axis) and absolute temperature, $T$ (along $x$-axis) for equal moles of two gases has been drawn. Given that volume of second gas is more than volume of first gas. Which of the following statement is correct?
1 Slope of gas 1 is less than gas 2
2 Slope of gas 1 is more than gas 2
3 Both have some slopes
4 None of the above
Explanation:
B According to an ideal gas equation- $\mathrm{PV}=\mathrm{nRT}$ $\frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{V}}$ $\frac{\mathrm{P}}{\mathrm{T}}$ represents slop of the graph as the number of moles are the same for the two gases. Then $\quad \frac{\mathrm{P}}{\mathrm{T}} \propto \frac{1}{\mathrm{~V}}$ Given, $\quad \mathrm{V}_{2}>\mathrm{V}_{1}$ slope $_{1}>$ slope $_{2}$
