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JEE physics MCQ

Thermodynamics · CBSE STD 12 Science (CBSE - English Medium). 783 questions.

Questions · Page 4 of 16

MCQ

MCQ 1511 Mark
A diatomic gas initially at $18^{\circ} \mathrm{C}$ is compressed adiabatically to oneeighth of its original volume. The temperature after compression will be
  • A
    $10^{\circ} \mathrm{C}$
  • B
    $887^{\circ} \mathrm{C}$
  • ✓
    $668 K$
  • D
    $144^{\circ} \mathrm{C}$
Answer
Correct option: C.
$668 K$
$T V^{\gamma-1}=\mathrm{constant}[\Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=(273+18)\left(\frac{V}{V} / {8}\right)^{0.4}=668 \mathrm{~K}]$
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MCQ 1521 Mark
A diatomic gas initially at $18^{\circ} \mathrm{C}$ is compressed adiabatically to oneeighth of its original volume. The temperature after compression will be
  • A
    $10^{\circ} \mathrm{C}$
  • B
    $887^{\circ} \mathrm{C}$
  • ✓
    $668 K$
  • D
    $144^{\circ} \mathrm{C}$
Answer
Correct option: C.
$668 K$
$T V^{\gamma-1}=\mathrm{constant} [\Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=(273+18)\left(\frac{V}{V / 8}\right)^{0.4}=668 \mathrm{~K}]$
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MCQ 1531 Mark
A diatomic gas initially at $18^{\circ} \mathrm{C}$ is compressed adiabatically to oneeighth of its original volume. The temperature after compression will be
  • A
    $10^{\circ} \mathrm{C}$
  • B
    $887^{\circ} \mathrm{C}$
  • ✓
    $668 K$
  • D
    $144^{\circ} \mathrm{C}$
Answer
Correct option: C.
$668 K$
$T V^{\gamma-1}=\mathrm{constant} [\Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=(273+18)\left(\frac{V}{V / 8}\right)^{0.4}=668 \mathrm{~K}]$
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MCQ 1541 Mark
In which thermodynamic process, volume remains same
  • A
    Isobaric
  • B
    Isothermal
  • C
    Adiabatic
  • ✓
    Isochoric
Answer
Correct option: D.
Isochoric
Isochoric
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MCQ 1551 Mark
A system is provided with $200 \mathrm{cal}$ of heat and the work done by the system on the surrounding is $40 \mathrm{~J}$. Then its internal energy
  • A
    Increases by $600 \mathrm{~J}$
  • B
    Decreases by $800 \mathrm{~J}$
  • ✓
    Increases by $800 \mathrm{~J}$
  • D
    Decreases by $50 \mathrm{~J}$
Answer
Correct option: C.
Increases by $800 \mathrm{~J}$
$\Delta Q=\Delta U+\Delta W $
$\because \Delta Q=200 \mathrm{cal}=200 \times 4.2=840 \mathrm{~J} \text { and } \Delta W=40 \mathrm{~J} $
$\Rightarrow \Delta U=\Delta Q-\Delta W$
$=840-40$
$=800 \mathrm{~J}$
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MCQ 1561 Mark
In which thermodynamic process, volume remains same
  • A
    Isobaric
  • B
    Isothermal
  • C
    Adiabatic
  • ✓
    Isochoric
Answer
Correct option: D.
Isochoric
Isochoric
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MCQ 1571 Mark
A system is provided with $200 \mathrm{cal}$ of heat and the work done by the system on the surrounding is $40 \mathrm{~J}$. Then its internal energy
  • A
    Increases by $600 \mathrm{~J}$
  • B
    Decreases by $800 \mathrm{~J}$
  • ✓
    Increases by $800 \mathrm{~J}$
  • D
    Decreases by $50 \mathrm{~J}$
Answer
Correct option: C.
Increases by $800 \mathrm{~J}$
$\Delta Q=\Delta U+\Delta W $
$\because \Delta Q=200 \mathrm{cal}=200 \times 4.2=840 \mathrm{~J} \text { and } \Delta W=40 \mathrm{~J} $
$\Rightarrow \Delta U=\Delta Q-\Delta W$
$=840-40$
$=800 \mathrm{~J}$
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MCQ 1581 Mark
In which thermodynamic process, volume remains same
  • A
    Isobaric
  • B
    Isothermal
  • C
    Adiabatic
  • ✓
    Isochoric
Answer
Correct option: D.
Isochoric
Isochoric
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MCQ 1591 Mark
A system is provided with $200 \mathrm{cal}$ of heat and the work done by the system on the surrounding is $40 \mathrm{~J}$. Then its internal energy
  • A
    Increases by $600 \mathrm{~J}$
  • B
    Decreases by $800 \mathrm{~J}$
  • ✓
    Increases by $800 \mathrm{~J}$
  • D
    Decreases by $50 \mathrm{~J}$
Answer
Correct option: C.
Increases by $800 \mathrm{~J}$
$\Delta Q=\Delta U+\Delta W $
$\because \Delta Q=200 \mathrm{cal}=200 \times 4.2=840 \mathrm{~J} \text { and } \Delta W=40 \mathrm{~J} $
$\Rightarrow \Delta U=\Delta Q-\Delta W$
$=840-40$
$=800 \mathrm{~J}$
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MCQ 1601 Mark
In an adiabatic process, the state of a gas is changed from $P_1, V_1, T_1$, to $P_2, V_2, T_2$. Which of the following relation is correct
  • ✓
    $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
  • B
    $P_1 V_1^{\gamma-1}=P_2 V_2^{\gamma-1}$
  • C
    $T_1 P_1^\gamma=T_2 P_2^\gamma$
  • D
    $T_1 V_1^\gamma=T_2 V_2^\gamma$
Answer
Correct option: A.
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
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MCQ 1611 Mark
In an adiabatic process, the state of a gas is changed from $P_1, V_1, T_1$, to $P_2, V_2, T_2$. Which of the following relation is correct
  • ✓
    $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
  • B
    $P_1 V_1^{\gamma-1}=P_2 V_2^{\gamma-1}$
  • C
    $T_1 P_1^\gamma=T_2 P_2^\gamma$
  • D
    $T_1 V_1^\gamma=T_2 V_2^\gamma$
Answer
Correct option: A.
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
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MCQ 1621 Mark
In an adiabatic process, the state of a gas is changed from $P_1, V_1, T_1$, to $P_2, V_2, T_2$. Which of the following relation is correct
  • ✓
    $T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
  • B
    $P_1 V_1^{\gamma-1}=P_2 V_2^{\gamma-1}$
  • C
    $T_1 P_1^\gamma=T_2 P_2^\gamma$
  • D
    $T_1 V_1^\gamma=T_2 V_2^\gamma$
Answer
Correct option: A.
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
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MCQ 1631 Mark
A thermodynamic system is taken through the cycle $\text{PQRSP}$ process. The net work done by the system isImage
  • ✓
    $20 \mathrm{~J}$
  • B
    $-20 \mathrm{~J}$
  • C
    $400 \mathrm{~J}$
  • D
    $-374 \mathrm{~J}$
Answer
Correct option: A.
$20 \mathrm{~J}$
Work done by the system $=$ Area of shaded portion on $P-V$ diagram
$[=(300-100)10^{-6} \times(200-10) \times 10^3=20 \mathrm{~J}]$
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MCQ 1641 Mark
A thermodynamic system is taken through the cycle $\text{PQRSP}$ process. The net work done by the system isImage
  • ✓
    $20 \mathrm{~J}$
  • B
    $-20 \mathrm{~J}$
  • C
    $400 \mathrm{~J}$
  • D
    $-374 \mathrm{~J}$
Answer
Correct option: A.
$20 \mathrm{~J}$
Work done by the system $=$ Area of shaded portion on $P-V$ diagram $[=(300-100)10^{-6} \times(200-10) \times 10^3=20 \mathrm{~J}]$
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MCQ 1651 Mark
A thermodynamic system is taken through the cycle $\text{PQRSP}$ process. The net work done by the system isImage
  • ✓
    $20 \mathrm{~J}$
  • B
    $-20 \mathrm{~J}$
  • C
    $400 \mathrm{~J}$
  • D
    $-374 \mathrm{~J}$
Answer
Correct option: A.
$20 \mathrm{~J}$
Work done by the system $=$ Area of shaded portion on $P-V$ diagram
$[=(300-100)10^{-6} \times(200-10) \times 10^3=20 \mathrm{~J}]$
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MCQ 1661 Mark
If $\gamma$ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal $P V$ curves at their point of intersection is
  • A
    $1 / \gamma$
  • ✓
    $\gamma$
  • C
    $\gamma-1$
  • D
    $\gamma+1$
Answer
Correct option: B.
$\gamma$
Slope of adiabatic curve $=\gamma \times($ Slope of isothermal curve $)$
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MCQ 1671 Mark
If $\gamma$ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal $P V$ curves at their point of intersection is
  • A
    $1 / \gamma$
  • ✓
    $\gamma$
  • C
    $\gamma-1$
  • D
    $\gamma+1$
Answer
Correct option: B.
$\gamma$
Slope of adiabatic curve $=\gamma \times($ Slope of isothermal curve $)$
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MCQ 1681 Mark
If $\gamma$ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal $P V$ curves at their point of intersection is
  • A
    $1 / \gamma$
  • ✓
    $\gamma$
  • C
    $\gamma-1$
  • D
    $\gamma+1$
Answer
Correct option: B.
$\gamma$
Slope of adiabatic curve $=\gamma \times($ Slope of isothermal curve $)$
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MCQ 1691 Mark
In a reversible isochoric change
  • ✓
    $\Delta W=0$
  • B
    $\Delta Q=0$
  • C
    $\Delta T=0$
  • D
    $\Delta U=0$
Answer
Correct option: A.
$\Delta W=0$
$\Delta V=0$
$\Rightarrow P \Delta V=0$
$\Rightarrow \Delta W=0$
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MCQ 1701 Mark
In a reversible isochoric change
  • ✓
    $\Delta W=0$
  • B
    $\Delta Q=0$
  • C
    $\Delta T=0$
  • D
    $\Delta U=0$
Answer
Correct option: A.
$\Delta W=0$
$\Delta V=0$
$\Rightarrow P \Delta V=0$
$\Rightarrow \Delta W=0$
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MCQ 1711 Mark
In a reversible isochoric change
  • ✓
    $\Delta W=0$
  • B
    $\Delta Q=0$
  • C
    $\Delta T=0$
  • D
    $\Delta U=0$
Answer
Correct option: A.
$\Delta W=0$
$\Delta V=0$
$\Rightarrow P \Delta V=0$
$\Rightarrow \Delta W=0$
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MCQ 1721 Mark
Air in a cylinder is suddenly compressed by a piston, which is then maintained at the same position. With the passage of time
  • ✓
    The pressure decreases
  • B
    The pressure increases
  • C
    The pressure remains the same
  • D
    The pressure may increase or decrease depending upon the nature of the gas
Answer
Correct option: A.
The pressure decreases
Due to compression the temperature of the system increases to a very highvalue.Thiscauses the flow of heat from system to the surroundings, thus decreasing thetemperature.Thisdecrease in temperature results in decrease in pressure.
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MCQ 1731 Mark
Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside
  • A
    Starts becoming hotter
  • B
    Remains at the same temperature
  • ✓
    Starts becoming cooler
  • D
    May become hotter or cooler depending upon the amount of water vapour present
Answer
Correct option: C.
Starts becoming cooler
Pressure is reduced, so the temperature falls.
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MCQ 1741 Mark
Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside
  • A
    Starts becoming hotter
  • B
    Remains at the same temperature
  • ✓
    Starts becoming cooler
  • D
    May become hotter or cooler depending upon the amount of water vapour present
Answer
Correct option: C.
Starts becoming cooler
Pressure is reduced, so the temperature falls.
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MCQ 1751 Mark
Compressed air in the tube of a wheel of a cycle at normal temperature suddenly starts coming out from a puncture. The air inside
  • A
    Starts becoming hotter
  • B
    Remains at the same temperature
  • ✓
    Starts becoming cooler
  • D
    May become hotter or cooler depending upon the amount of water vapour present
Answer
Correct option: C.
Starts becoming cooler
Pressure is reduced, so the temperature falls.
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MCQ 1761 Mark
In an isothermal process the volume of an ideal gas is halved. One can say that
  • A
    Internal energy of the system decreases
  • B
    Work done by the gas is positive
  • ✓
    Work done by the gas is negative
  • D
    Internal energy of the system increases
Answer
Correct option: C.
Work done by the gas is negative
(c) For isothermal process
$d U=0 \text { and work done }=d W=P\left(V_2-V_1\right) $
$\because V_2=\frac{V_1}{2}=\frac{V}{2} $
$\therefore d W=-\frac{P V}{2}$
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MCQ 1771 Mark
In an isothermal process the volume of an ideal gas is halved. One can say that
  • A
    Internal energy of the system decreases
  • B
    Work done by the gas is positive
  • ✓
    Work done by the gas is negative
  • D
    Internal energy of the system increases
Answer
Correct option: C.
Work done by the gas is negative
(c) For isothermal process
$d U=0 \text { and work done }=d W=P\left(V_2-V_1\right) $
$\because V_2=\frac{V_1}{2}=\frac{V}{2} $
$\therefore d W=-\frac{P V}{2}$
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MCQ 1781 Mark
In an isothermal process the volume of an ideal gas is halved. One can say that
  • A
    Internal energy of the system decreases
  • B
    Work done by the gas is positive
  • ✓
    Work done by the gas is negative
  • D
    Internal energy of the system increases
Answer
Correct option: C.
Work done by the gas is negative
(c) For isothermal process
$d U=0 \text { and work done }=d W=P\left(V_2-V_1\right) $
$\because V_2=\frac{V_1}{2}=\frac{V}{2} \therefore d W=-\frac{P V}{2}$
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MCQ 1791 Mark
In a thermodynamic system working substance is ideal gas, its internal energy is in the form of
  • ✓
    Kinetic energy only
  • B
    Kinetic and potential energy
  • C
    Potential energy
  • D
    None of these
Answer
Correct option: A.
Kinetic energy only
(a) Ideal gas possess only kinetic energy.
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MCQ 1801 Mark
In a thermodynamic system working substance is ideal gas, its internal energy is in the form of
  • ✓
    Kinetic energy only
  • B
    Kinetic and potential energy
  • C
    Potential energy
  • D
    None of these
Answer
Correct option: A.
Kinetic energy only
(a) Ideal gas possess only kinetic energy.
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MCQ 1811 Mark
In a thermodynamic system working substance is ideal gas, its internal energy is in the form of
  • ✓
    Kinetic energy only
  • B
    Kinetic and potential energy
  • C
    Potential energy
  • D
    None of these
Answer
Correct option: A.
Kinetic energy only
(a) Ideal gas possess only kinetic energy.
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MCQ 1821 Mark
Air is filled in a motor tube at $27^{\circ} \mathrm{C}$ and at a pressure of 8 atmospheres. The tube suddenly bursts, then temperature of air is [Given $\gamma$ of air $=1.5$ ]
  • A
    $27.5^{\circ} \mathrm{C}$
  • B
    $75^{\circ} K$
  • ✓
    $150 K$
  • D
    $150^{\circ} \mathrm{C}$
Answer
Correct option: C.
$150 K$
(c)
$\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} $
$\Rightarrow \frac{T_2}{T_1}=\left(\frac{1}{8}\right)^{\frac{1.5-1}{1.5}}=\left(\frac{1}{8}\right)^{\frac{1}{3}}=\frac{1}{2} $
$\Rightarrow T_2=\frac{T_1}{2}=\frac{300}{2}=150 \mathrm{~K} .$
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MCQ 1831 Mark
Air is filled in a motor tube at $27^{\circ} \mathrm{C}$ and at a pressure of 8 atmospheres. The tube suddenly bursts, then temperature of air is [Given $\gamma$ of air $=1.5$ ]
  • A
    $27.5^{\circ} \mathrm{C}$
  • B
    $75^{\circ} K$
  • ✓
    $150 K$
  • D
    $150^{\circ} \mathrm{C}$
Answer
Correct option: C.
$150 K$
(c)
$\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} $
$\Rightarrow \frac{T_2}{T_1}=\left(\frac{1}{8}\right)^{\frac{1.5-1}{1.5}}=\left(\frac{1}{8}\right)^{\frac{1}{3}}=\frac{1}{2} $
$\Rightarrow T_2=\frac{T_1}{2}=\frac{300}{2}=150 \mathrm{~K} .$
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MCQ 1841 Mark
Air is filled in a motor tube at $27^{\circ} \mathrm{C}$ and at a pressure of 8 atmospheres. The tube suddenly bursts, then temperature of air is [Given $\gamma$ of air $=1.5$ ]
  • A
    $27.5^{\circ} \mathrm{C}$
  • B
    $75^{\circ} K$
  • ✓
    $150 K$
  • D
    $150^{\circ} \mathrm{C}$
Answer
Correct option: C.
$150 K$
(c)
$\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} $
$\Rightarrow \frac{T_2}{T_1}=\left(\frac{1}{8}\right)^{\frac{1.5-1}{1.5}}=\left(\frac{1}{8}\right)^{\frac{1}{3}}=\frac{1}{2} $
$\Rightarrow T_2=\frac{T_1}{2}=\frac{300}{2}=150 \mathrm{~K} .$
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MCQ 1851 Mark
In the cyclic process shown in the figure, the work done by the gas in one cycle is
Image
  • A
    $28 P_1 V_1$
  • B
    $14 P_1 V_1$
  • C
    $18 P_1 V_1$
  • ✓
    $9 P_1 V_1$
Answer
Correct option: D.
$9 P_1 V_1$
(d) Work done $=$ Area under curve $=\frac{6 P_1 \times 3 V_1}{2}=9 \mathrm{PV}$
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MCQ 1861 Mark
In the cyclic process shown in the figure, the work done by the gas in one cycle is
Image
  • A
    $28 P_1 V_1$
  • B
    $14 P_1 V_1$
  • C
    $18 P_1 V_1$
  • ✓
    $9 P_1 V_1$
Answer
Correct option: D.
$9 P_1 V_1$
(d) Work done $=$ Area under curve $=\frac{6 P_1 \times 3 V_1}{2}=9 \mathrm{PV}$
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MCQ 1871 Mark
In the cyclic process shown in the figure, the work done by the gas in one cycle is
Image
  • A
    $28 P_1 V_1$
  • B
    $14 P_1 V_1$
  • C
    $18 P_1 V_1$
  • ✓
    $9 P_1 V_1$
Answer
Correct option: D.
$9 P_1 V_1$
(d) Work done $=$ Area under curve $=\frac{6 P_1 \times 3 V_1}{2}=9 \mathrm{PV}$
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MCQ 1881 Mark
One mole of an ideal gas expands at a constant temperature of 300 $K$ from an initial volume of 10 litres to a final volume of 20 litres. The work done in expanding the gas is $(R=8.31 \mathrm{~J} / \mathrm{mole}-K)$
  • A
    750 joules
  • ✓
    1728 joules
  • C
    1500 joules
  • D
    3456 joules
Answer
Correct option: B.
1728 joules
(b)$W_{i s o}=\mu R T \log _e \frac{V_2}{V_1}=1 \times 8.31 \times 300 \log _e \frac{20}{10}=1728 \mathrm{~J}$
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MCQ 1891 Mark
One mole of an ideal gas expands at a constant temperature of 300 $K$ from an initial volume of 10 litres to a final volume of 20 litres. The work done in expanding the gas is $(R=8.31 \mathrm{~J} / \mathrm{mole}-K)$
  • A
    750 joules
  • ✓
    1728 joules
  • C
    1500 joules
  • D
    3456 joules
Answer
Correct option: B.
1728 joules
(b)$W_{i s o}=\mu R T \log _e \frac{V_2}{V_1}=1 \times 8.31 \times 300 \log _e \frac{20}{10}=1728 \mathrm{~J}$
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MCQ 1901 Mark
One mole of an ideal gas expands at a constant temperature of 300 $K$ from an initial volume of 10 litres to a final volume of 20 litres. The work done in expanding the gas is $(R=8.31 \mathrm{~J} / \mathrm{mole}-K)$
  • A
    750 joules
  • ✓
    1728 joules
  • C
    1500 joules
  • D
    3456 joules
Answer
Correct option: B.
1728 joules
(b)$W_{i s o}=\mu R T \log _e \frac{V_2}{V_1}=1 \times 8.31 \times 300 \log _e \frac{20}{10}=1728 \mathrm{~J}$
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MCQ 1911 Mark
An ideal gas is expanded adiabatically at an initial temperature of $300 K$ so that its volume is doubled. The final temperature of the hydrogen gas is $(\gamma=1.40)$
  • ✓
    $227.36 K$
  • B
    $500.30 \mathrm{~K}$
  • C
    $454.76 K$
  • D
    $-47^{\circ} \mathrm{C}$
Answer
Correct option: A.
$227.36 K$
(a)$T V^{\gamma-1}=\text {constant}$
$\Rightarrow\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^\gamma $
$\Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^\gamma $
$\Rightarrow T_2=300\left(\frac{1}{2}\right)^{0.4}=227.36 \mathrm{~K}$
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MCQ 1921 Mark
An ideal gas is expanded adiabatically at an initial temperature of $300 K$ so that its volume is doubled. The final temperature of the hydrogen gas is $(\gamma=1.40)$
  • ✓
    $227.36 K$
  • B
    $500.30 \mathrm{~K}$
  • C
    $454.76 K$
  • D
    $-47^{\circ} \mathrm{C}$
Answer
Correct option: A.
$227.36 K$
(a)
$T V^{\gamma-1}=\text {constant}$
$\Rightarrow\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^\gamma $
$\Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^\gamma $
$\Rightarrow T_2=300\left(\frac{1}{2}\right)^{0.4}=227.36 \mathrm{~K}$
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MCQ 1931 Mark
An ideal gas is expanded adiabatically at an initial temperature of $300 K$ so that its volume is doubled. The final temperature of the hydrogen gas is $(\gamma=1.40)$
  • ✓
    $227.36 K$
  • B
    $500.30 \mathrm{~K}$
  • C
    $454.76 K$
  • D
    $-47^{\circ} \mathrm{C}$
Answer
Correct option: A.
$227.36 K$
 
(a)
$T V^{\gamma-1}=\text {constant}$
$\Rightarrow\frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^\gamma $
$\Rightarrow T_2=T_1\left(\frac{V_1}{V_2}\right)^\gamma $
$\Rightarrow T_2=300\left(\frac{1}{2}\right)^{0.4}=227.36 \mathrm{~K}$
 
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MCQ 1941 Mark
Which is the correct statement
  • ✓
    For an isothermal change $P V=$ constant
  • B
    In an isothermal process the change in internal energy must be equal to the work done
  • C
    For an adiabatic change $\frac{P_2}{P_1}=\left(\frac{V_2}{V_1}\right)^\gamma$, where $\gamma$ is the ratio of specific heats
  • D
    In an adiabatic process work done must be equal to the heat entering the system
Answer
Correct option: A.
For an isothermal change $P V=$ constant
(a) Since $P V=R T$ and $T=$ constant; $\therefore P V=$ constant.
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MCQ 1951 Mark
Which is the correct statement
  • ✓
    For an isothermal change $P V=$ constant
  • B
    In an isothermal process the change in internal energy must be equal to the work done
  • C
    For an adiabatic change $\frac{P_2}{P_1}=\left(\frac{V_2}{V_1}\right)^\gamma$, where $\gamma$ is the ratio of specific heats
  • D
    In an adiabatic process work done must be equal to the heat entering the system
Answer
Correct option: A.
For an isothermal change $P V=$ constant
(a) Since $P V=R T$ and $T=$ constant; $\therefore P V=$ constant.
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MCQ 1961 Mark
Which is the correct statement
  • ✓
    For an isothermal change $P V=$ constant
  • B
    In an isothermal process the change in internal energy must be equal to the work done
  • C
    For an adiabatic change $\frac{P_2}{P_1}=\left(\frac{V_2}{V_1}\right)^\gamma$, where $\gamma$ is the ratio of specific heats
  • D
    In an adiabatic process work done must be equal to the heat entering the system
Answer
Correct option: A.
For an isothermal change $P V=$ constant
(a) Since $P V=R T$ and $T=$ constant; $\therefore P V=$ constant.
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MCQ 1971 Mark
A thermodynamic system goes from states (i) $P_1, V$ to $2 P_1, V$ (ii) $P, V$ to $P, 2 V$. Then work done in the two cases is
  • A
    Zero, Zero
  • ✓
    Zero, $P V_1$
  • C
    $P V_1$, Zero
  • D
    $P V_1, P_1 V_1$
Answer
Correct option: B.
Zero, $P V_1$
(i) Case $\rightarrow$ Volume $=$ constant $\Rightarrow \int P d V=0$
(ii) Case $\rightarrow P=$ constant $\Rightarrow \int_{V_1}^{2 V_1} P d V=P \int_{V_1}^{2 V_1} d V=P V_1$
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MCQ 1981 Mark
The temperature of an ideal gas is kept constant as it expands. The gas does external work. During this process, the internal energy of the gas
  • A
    Decreases
  • B
    Increases
  • ✓
    Remains constant
  • D
    Depends on the molecular motion
Answer
Correct option: C.
Remains constant
(c) Internal energy depends only on the temperature of the gas.
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MCQ 1991 Mark
The adiabatic elasticity of hydrogen gas $(\gamma=1.4)$ at NTP is
  • A
    $1 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
  • B
    $1 \times 10^{-8} \mathrm{~N} / \mathrm{m} 2$
  • C
    $1.4 \mathrm{~N} / \mathrm{m}^2$
  • ✓
    $1.4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
Answer
Correct option: D.
$1.4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
(d) $E_\phi=\gamma P=1.4 \times\left(1 \times 10^5\right)=1.4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
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MCQ 2001 Mark
A thermodynamic system goes from states (i) $P_1, V$ to $2 P_1, V$ (ii) $P, V$ to $P, 2 V$. Then work done in the two cases is
  • A
    Zero, Zero
  • ✓
    Zero, $P V_1$
  • C
    $P V_1$, Zero
  • D
    $P V_1, P_1 V_1$
Answer
Correct option: B.
Zero, $P V_1$
(i) Case $\rightarrow$ Volume $=$ constant $\Rightarrow \int P d V=0$
(ii) Case $\rightarrow P=$ constant $\Rightarrow \int_{V_1}^{2 V_1} P d V=P \int_{V_1}^{2 V_1} d V=P V_1$
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MCQ - Page 4 - JEE physics STD 12 Science Questions - Vidyadip