NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
148193
An ideal gas is taken through a cycle $\mathrm{ABCA}$ as shown below in P-V diagram. What is the work done during the cycle?
1 $P V / 2$
2 PV
3 2 PV
4 Zero
Explanation:
B Work done $=$ Area enclosed by triangle $\mathrm{ABCA}$ $=\frac{1}{2}(2 \mathrm{P}-\mathrm{P})(3 \mathrm{~V}-\mathrm{V})$ $=\frac{1}{2}(\mathrm{P})(2 \mathrm{~V})$ $\therefore \quad$ Work done $=\mathrm{PV}$
SCRA-2012
Thermodynamics
148195
Helium gas expands at a constant pressure when $15 \mathrm{~kJ}$ of heat is supplied. If $C_{p}: C_{v}$ is equal to $5: 3$, what is the increase in the internal energy?
1 $3 \mathrm{~kJ}$
2 $6 \mathrm{~kJ}$
3 $9 \mathrm{~kJ}$
4 $15 \mathrm{~kJ}$
Explanation:
C Given that, $\mathrm{Q}=15 \mathrm{~kJ}, \quad \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}$ According to the first law of thermodynamics - $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{U}}{\mathrm{Q}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}$ $\Delta \mathrm{U}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \times \mathrm{Q}=\frac{3}{5} \times 15 \mathrm{~kJ}$ $\Delta \mathrm{U}=9 \mathrm{~kJ}$
SCRA-2010
Thermodynamics
148200
If for a gas $R / C_{V}=0.67$, this gas is made up of molecules which are
1 monoatomic
2 diatomic
3 polyatomic
4 mixture of diatomic and polyatomic molecules
Explanation:
A Given that, $\mathrm{R} / \mathrm{C}_{\mathrm{V}}=0.67$ We know that, $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ $\therefore \quad \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}=0.67$ or $\quad \frac{C_{p}}{C_{v}}-1=0.670$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.67$ So, the gas is monoatomic.
J and K CET- 2003
Thermodynamics
148204
An ideal gas undergoes through an isochoric process whereby its internal energy is increased by $\Delta \mathrm{U}$. If the heat expelled (thrown out) by the gas is $\Delta \mathrm{Q}$, then the ratio $\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}$ is
1 1.0
2 0.5
3 -1.0
4 -2.0
Explanation:
C Internal energy increases means temperature increases and work done be zero as it is isochoric process. Given heat expelled by the gas so $\Delta \mathrm{Q}$ must be negative. $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=-\Delta \mathrm{U}$ $\therefore \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=-1$
148193
An ideal gas is taken through a cycle $\mathrm{ABCA}$ as shown below in P-V diagram. What is the work done during the cycle?
1 $P V / 2$
2 PV
3 2 PV
4 Zero
Explanation:
B Work done $=$ Area enclosed by triangle $\mathrm{ABCA}$ $=\frac{1}{2}(2 \mathrm{P}-\mathrm{P})(3 \mathrm{~V}-\mathrm{V})$ $=\frac{1}{2}(\mathrm{P})(2 \mathrm{~V})$ $\therefore \quad$ Work done $=\mathrm{PV}$
SCRA-2012
Thermodynamics
148195
Helium gas expands at a constant pressure when $15 \mathrm{~kJ}$ of heat is supplied. If $C_{p}: C_{v}$ is equal to $5: 3$, what is the increase in the internal energy?
1 $3 \mathrm{~kJ}$
2 $6 \mathrm{~kJ}$
3 $9 \mathrm{~kJ}$
4 $15 \mathrm{~kJ}$
Explanation:
C Given that, $\mathrm{Q}=15 \mathrm{~kJ}, \quad \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}$ According to the first law of thermodynamics - $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{U}}{\mathrm{Q}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}$ $\Delta \mathrm{U}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \times \mathrm{Q}=\frac{3}{5} \times 15 \mathrm{~kJ}$ $\Delta \mathrm{U}=9 \mathrm{~kJ}$
SCRA-2010
Thermodynamics
148200
If for a gas $R / C_{V}=0.67$, this gas is made up of molecules which are
1 monoatomic
2 diatomic
3 polyatomic
4 mixture of diatomic and polyatomic molecules
Explanation:
A Given that, $\mathrm{R} / \mathrm{C}_{\mathrm{V}}=0.67$ We know that, $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ $\therefore \quad \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}=0.67$ or $\quad \frac{C_{p}}{C_{v}}-1=0.670$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.67$ So, the gas is monoatomic.
J and K CET- 2003
Thermodynamics
148204
An ideal gas undergoes through an isochoric process whereby its internal energy is increased by $\Delta \mathrm{U}$. If the heat expelled (thrown out) by the gas is $\Delta \mathrm{Q}$, then the ratio $\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}$ is
1 1.0
2 0.5
3 -1.0
4 -2.0
Explanation:
C Internal energy increases means temperature increases and work done be zero as it is isochoric process. Given heat expelled by the gas so $\Delta \mathrm{Q}$ must be negative. $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=-\Delta \mathrm{U}$ $\therefore \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=-1$
148193
An ideal gas is taken through a cycle $\mathrm{ABCA}$ as shown below in P-V diagram. What is the work done during the cycle?
1 $P V / 2$
2 PV
3 2 PV
4 Zero
Explanation:
B Work done $=$ Area enclosed by triangle $\mathrm{ABCA}$ $=\frac{1}{2}(2 \mathrm{P}-\mathrm{P})(3 \mathrm{~V}-\mathrm{V})$ $=\frac{1}{2}(\mathrm{P})(2 \mathrm{~V})$ $\therefore \quad$ Work done $=\mathrm{PV}$
SCRA-2012
Thermodynamics
148195
Helium gas expands at a constant pressure when $15 \mathrm{~kJ}$ of heat is supplied. If $C_{p}: C_{v}$ is equal to $5: 3$, what is the increase in the internal energy?
1 $3 \mathrm{~kJ}$
2 $6 \mathrm{~kJ}$
3 $9 \mathrm{~kJ}$
4 $15 \mathrm{~kJ}$
Explanation:
C Given that, $\mathrm{Q}=15 \mathrm{~kJ}, \quad \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}$ According to the first law of thermodynamics - $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{U}}{\mathrm{Q}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}$ $\Delta \mathrm{U}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \times \mathrm{Q}=\frac{3}{5} \times 15 \mathrm{~kJ}$ $\Delta \mathrm{U}=9 \mathrm{~kJ}$
SCRA-2010
Thermodynamics
148200
If for a gas $R / C_{V}=0.67$, this gas is made up of molecules which are
1 monoatomic
2 diatomic
3 polyatomic
4 mixture of diatomic and polyatomic molecules
Explanation:
A Given that, $\mathrm{R} / \mathrm{C}_{\mathrm{V}}=0.67$ We know that, $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ $\therefore \quad \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}=0.67$ or $\quad \frac{C_{p}}{C_{v}}-1=0.670$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.67$ So, the gas is monoatomic.
J and K CET- 2003
Thermodynamics
148204
An ideal gas undergoes through an isochoric process whereby its internal energy is increased by $\Delta \mathrm{U}$. If the heat expelled (thrown out) by the gas is $\Delta \mathrm{Q}$, then the ratio $\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}$ is
1 1.0
2 0.5
3 -1.0
4 -2.0
Explanation:
C Internal energy increases means temperature increases and work done be zero as it is isochoric process. Given heat expelled by the gas so $\Delta \mathrm{Q}$ must be negative. $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=-\Delta \mathrm{U}$ $\therefore \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=-1$
148193
An ideal gas is taken through a cycle $\mathrm{ABCA}$ as shown below in P-V diagram. What is the work done during the cycle?
1 $P V / 2$
2 PV
3 2 PV
4 Zero
Explanation:
B Work done $=$ Area enclosed by triangle $\mathrm{ABCA}$ $=\frac{1}{2}(2 \mathrm{P}-\mathrm{P})(3 \mathrm{~V}-\mathrm{V})$ $=\frac{1}{2}(\mathrm{P})(2 \mathrm{~V})$ $\therefore \quad$ Work done $=\mathrm{PV}$
SCRA-2012
Thermodynamics
148195
Helium gas expands at a constant pressure when $15 \mathrm{~kJ}$ of heat is supplied. If $C_{p}: C_{v}$ is equal to $5: 3$, what is the increase in the internal energy?
1 $3 \mathrm{~kJ}$
2 $6 \mathrm{~kJ}$
3 $9 \mathrm{~kJ}$
4 $15 \mathrm{~kJ}$
Explanation:
C Given that, $\mathrm{Q}=15 \mathrm{~kJ}, \quad \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{5}{3}$ According to the first law of thermodynamics - $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}$ $\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ $\frac{\Delta \mathrm{U}}{\mathrm{Q}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}$ $\Delta \mathrm{U}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}} \times \mathrm{Q}=\frac{3}{5} \times 15 \mathrm{~kJ}$ $\Delta \mathrm{U}=9 \mathrm{~kJ}$
SCRA-2010
Thermodynamics
148200
If for a gas $R / C_{V}=0.67$, this gas is made up of molecules which are
1 monoatomic
2 diatomic
3 polyatomic
4 mixture of diatomic and polyatomic molecules
Explanation:
A Given that, $\mathrm{R} / \mathrm{C}_{\mathrm{V}}=0.67$ We know that, $\mathrm{R}=\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}$ $\therefore \quad \frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\frac{\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}}{\mathrm{C}_{\mathrm{v}}}=0.67$ or $\quad \frac{C_{p}}{C_{v}}-1=0.670$ $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.67$ So, the gas is monoatomic.
J and K CET- 2003
Thermodynamics
148204
An ideal gas undergoes through an isochoric process whereby its internal energy is increased by $\Delta \mathrm{U}$. If the heat expelled (thrown out) by the gas is $\Delta \mathrm{Q}$, then the ratio $\frac{\Delta \mathrm{U}}{\Delta \mathrm{Q}}$ is
1 1.0
2 0.5
3 -1.0
4 -2.0
Explanation:
C Internal energy increases means temperature increases and work done be zero as it is isochoric process. Given heat expelled by the gas so $\Delta \mathrm{Q}$ must be negative. $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\Delta \mathrm{Q}=-\Delta \mathrm{U}$ $\therefore \frac{\Delta \mathrm{Q}}{\Delta \mathrm{U}}=-1$
