Download App

JEE physics MCQ

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

Questions · Page 5 of 16

MCQ

MCQ 2011 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.
View full question & answer→
MCQ 2021 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) $E_\phi=\gamma P=1.4 \times\left(1 \times 10^5\right)=1.4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
View full question & answer→
MCQ 2031 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 $\Rightarrow \int_{V_1}^{2 V_1} P d V=P \int_{V_1}^{2 V_1} d V=P V_1$
View full question & answer→
MCQ 2041 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.
View full question & answer→
MCQ 2051 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) $E_\phi=\gamma P=1.4 \times\left(1 \times 10^5\right)=1.4 \times 10^5 \mathrm{~N} / \mathrm{m}^2$
View full question & answer→
MCQ 2061 Mark
The internal energy of the gas increases $\ln$
  • A
    Adiabatic expansion
  • ✓
    Adiabatic compression
  • C
    lsothermal expansion
  • D
    Isothermal compression
Answer
Correct option: B.
Adiabatic compression
(b) In adiabatic process $\Delta U=-\Delta W$. In compression $\Delta W$ is negative, so $\Delta U$ is positive i.e. internal energy increases.
View full question & answer→
MCQ 2071 Mark
The internal energy of the gas increases $\ln$
  • A
    Adiabatic expansion
  • ✓
    Adiabatic compression
  • C
    lsothermal expansion
  • D
    Isothermal compression
Answer
Correct option: B.
Adiabatic compression
(b) In adiabatic process $\Delta U=-\Delta W$. In compression $\Delta W$ is negative, so $\Delta U$ is positive i.e. internal energy increases.
View full question & answer→
MCQ 2081 Mark
The internal energy of the gas increases $\ln$
  • A
    Adiabatic expansion
  • ✓
    Adiabatic compression
  • C
    lsothermal expansion
  • D
    Isothermal compression
Answer
Correct option: B.
Adiabatic compression
(b) In adiabatic process $\Delta U=-\Delta W$. In compression $\Delta W$ is negative, so $\Delta U$ is positive i.e. internal energy increases.
View full question & answer→
MCQ 2091 Mark
If the amount of heat given to a system be 35 joules and the amount of work done by the system be -15 joules, then the change in the internal energy of the system is
  • A
    -50 joules
  • B
    20 joules
  • C
    30 joules
  • ✓
    50 joules
Answer
Correct option: D.
50 joules
(d) $\Delta Q=\Delta W+\Delta U \Rightarrow 35=-15+\Delta U \Rightarrow \Delta U=50 J$
View full question & answer→
MCQ 2101 Mark
The amount of work done in an adiabatic expansion from temperature $T$ to $T_1$ is
  • A
    $R\left(T-T_1\right)$
  • ✓
    $\frac{R}{\gamma-1}\left(T-T_1\right)$
  • C
    $R T$
  • D
    $R\left(T-T_1\right)(\gamma-1)$
Answer
Correct option: B.
$\frac{R}{\gamma-1}\left(T-T_1\right)$
(b) $W_{a d i}=\frac{R}{\gamma-1}\left(T_i-T_f\right)=\frac{R}{\gamma-1}\left(T-T_1\right)$
View full question & answer→
MCQ 2111 Mark
If the amount of heat given to a system be $35$ joules and the amount of work done by the system be $-15$ joules, then the change in the internal energy of the system is
  • A
    $-50$ joules
  • B
    $20$ joules
  • C
    $30$ joules
  • ✓
    $50$ joules
Answer
Correct option: D.
$50$ joules
$\Delta Q=\Delta W+\Delta U \Rightarrow 35=-15+\Delta U \Rightarrow \Delta U=50 J$
View full question & answer→
MCQ 2121 Mark
The amount of work done in an adiabatic expansion from temperature $T$ to $T_1$ is
  • A
    $R\left(T-T_1\right)$
  • ✓
    $\frac{R}{\gamma-1}\left(T-T_1\right)$
  • C
    $R T$
  • D
    $R\left(T-T_1\right)(\gamma-1)$
Answer
Correct option: B.
$\frac{R}{\gamma-1}\left(T-T_1\right)$
$W_{a d i}=\frac{R}{\gamma-1}\left(T_i-T_f\right)=\frac{R}{\gamma-1}\left(T-T_1\right)$
View full question & answer→
MCQ 2131 Mark
If the amount of heat given to a system be $35$ joules and the amount of work done by the system be $-15$ joules, then the change in the internal energy of the system is
  • A
    $-50$ joules
  • B
    $20$ joules
  • C
    $30$ joules
  • ✓
    $50$ joules
Answer
Correct option: D.
$50$ joules
$\Delta Q=\Delta W+\Delta U$
$\Rightarrow 35=-15+\Delta U$
$\Rightarrow \Delta U=50 J$
View full question & answer→
MCQ 2141 Mark
The amount of work done in an adiabatic expansion from temperature $T$ to $T_1$ is
  • A
    $R\left(T-T_1\right)$
  • ✓
    $\frac{R}{\gamma-1}\left(T-T_1\right)$
  • C
    $R T$
  • D
    $R\left(T-T_1\right)(\gamma-1)$
Answer
Correct option: B.
$\frac{R}{\gamma-1}\left(T-T_1\right)$
$W_{a d i}=\frac{R}{\gamma-1}\left(T_i-T_f\right)=\frac{R}{\gamma-1}\left(T-T_1\right)$
View full question & answer→
MCQ 2151 Mark
The first law of thermodynamics is concerned with the conservation of
  • A
    Momentum
  • ✓
    Energy
  • C
    Mass
  • D
    Temperature
Answer
Correct option: B.
Energy
Energy
View full question & answer→
MCQ 2161 Mark
The first law of thermodynamics is concerned with the conservation of
  • A
    Momentum
  • ✓
    Energy
  • C
    Mass
  • D
    Temperature
Answer
Correct option: B.
Energy
Energy
View full question & answer→
MCQ 2171 Mark
The first law of thermodynamics is concerned with the conservation of
  • A
    Momentum
  • ✓
    Energy
  • C
    Mass
  • D
    Temperature
Answer
Correct option: B.
Energy
Energy
View full question & answer→
MCQ 2181 Mark
When an ideal gas in a cylinder was compressed isothermally by a piston, the work done on the gas was found to be $1.5 \times 10^4$ joules. During this process about
  • ✓
    $3.6 \times 10^3 \mathrm{cal}$ of heat flowed out from the gas
  • B
    $3.6 \times 10^3 \mathrm{cal}$ of heat flowed into the gas
  • C
    $1.5 \times 10^4 \mathrm{cal}$ of heat flowed into the gas
  • D
    $1.5 \times 10^4 \mathrm{cal}$ of heat flowed out from the gas
Answer
Correct option: A.
$3.6 \times 10^3 \mathrm{cal}$ of heat flowed out from the gas
In isothermal compression, there is always an increase of heat. which must flow out the gas.
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta W$
$(\because\Delta U=0)$
$\Rightarrow \Delta Q=-1.5 \times 10^4\mathrm{~J}=\frac{1.5\times10^4}{4.18}\mathrm{cal}=-3.6\times10^3\mathrm{cal}$
View full question & answer→
MCQ 2191 Mark
When an ideal gas in a cylinder was compressed isothermally by a piston, the work done on the gas was found to be $1.5 \times 10^4$ joules. During this process about
  • ✓
    $3.6 \times 10^3 \mathrm{cal}$ of heat flowed out from the gas
  • B
    $3.6 \times 10^3 \mathrm{cal}$ of heat flowed into the gas
  • C
    $1.5 \times 10^4 \mathrm{cal}$ of heat flowed into the gas
  • D
    $1.5 \times 10^4 \mathrm{cal}$ of heat flowed out from the gas
Answer
Correct option: A.
$3.6 \times 10^3 \mathrm{cal}$ of heat flowed out from the gas
In isothermal compression, there is always an increase of heat. which must flow out the gas.
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta W$
$(\because\Delta U=0)$
$\Rightarrow \Delta Q=-1.5 \times 10^4\mathrm{~J}=\frac{1.5\times10^4}{4.18}\mathrm{cal}=-3.6\times10^3\mathrm{cal}$
View full question & answer→
MCQ 2201 Mark
When an ideal gas in a cylinder was compressed isothermally by a piston, the work done on the gas was found to be $1.5 \times 10^4$ joules. During this process about
  • ✓
    $3.6 \times 10^3 \mathrm{cal}$ of heat flowed out from the gas
  • B
    $3.6 \times 10^3 \mathrm{cal}$ of heat flowed into the gas
  • C
    $1.5 \times 10^4 \mathrm{cal}$ of heat flowed into the gas
  • D
    $1.5 \times 10^4 \mathrm{cal}$ of heat flowed out from the gas
Answer
Correct option: A.
$3.6 \times 10^3 \mathrm{cal}$ of heat flowed out from the gas
In isothermal compression, there is always an increase of heat. which must flow out the gas.
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta W$
$(\because\Delta U=0)$
$\Rightarrow \Delta Q=-1.5 \times 10^4\mathrm{~J}=\frac{1.5\times10^4}{4.18}\mathrm{cal}=-3.6\times10^3\mathrm{cal}$
 
View full question & answer→
MCQ 2211 Mark
In changing the state of thermodynamics from $A$ to $B$ state, the heat required is $Q$ and the work done by the system is $\mathrm{W}$. The change in its internal energy is
  • A
    $Q+W$
  • ✓
    $Q-W$
  • C
    $Q$
  • D
    $\frac{Q-W}{2}$
Answer
Correct option: B.
$Q-W$
$\Delta Q=\Delta U+\Delta W $
$\Rightarrow \Delta U=\Delta Q-\Delta W=Q-W \text { (using proper sign) }$
View full question & answer→
MCQ 2221 Mark
In changing the state of thermodynamics from $A$ to $B$ state, the heat required is $Q$ and the work done by the system is $\mathrm{W}$. The change in its internal energy is
  • A
    $Q+W$
  • ✓
    $Q-W$
  • C
    $Q$
  • D
    $\frac{Q-W}{2}$
Answer
Correct option: B.
$Q-W$
$\Delta Q=\Delta U+\Delta W $
$\Rightarrow \Delta U=\Delta Q-\Delta W=Q-W \text { (using proper sign) }$
View full question & answer→
MCQ 2231 Mark
In changing the state of thermodynamics from $A$ to $B$ state, the heat required is $Q$ and the work done by the system is $\mathrm{W}$. The change in its internal energy is
  • A
    $Q+W$
  • ✓
    $Q-W$
  • C
    $Q$
  • D
    $\frac{Q-W}{2}$
Answer
Correct option: B.
$Q-W$
$\Delta Q=\Delta U+\Delta W $
$\Rightarrow \Delta U=\Delta Q-\Delta W=Q-W \text { (using proper sign) }$
View full question & answer→
MCQ 2241 Mark
Heat given to a system is $35$ joules and work done by the system is $15$ joules. The change in the internal energy of the system will be
  • A
    $-50 \mathrm{~J}$
  • ✓
    $20 \mathrm{~J}$
  • C
    $30 \mathrm{~J}$
  • D
    $50 \mathrm{~J}$
Answer
Correct option: B.
$20 \mathrm{~J}$
$\Delta U=\Delta Q-W=35-15=20 \mathrm{~J}$
View full question & answer→
MCQ 2251 Mark
If a cylinder containing a gas at high pressure explodes, the gas undergoes
  • A
    Reversible adiabatic change and fall of temperature
  • B
    Reversible adiabatic change and rise of temperature
  • ✓
    Irreversible adiabatic change and fall of temperature
  • D
    Irreversible adiabatic change and rise of temperature
Answer
Correct option: C.
Irreversible adiabatic change and fall of temperature
Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature.
View full question & answer→
MCQ 2261 Mark
Heat given to a system is $35$ joules and work done by the system is $15$ joules. The change in the internal energy of the system will be
  • A
    $-50 \mathrm{~J}$
  • ✓
    $20 \mathrm{~J}$
  • C
    $30 \mathrm{~J}$
  • D
    $50 \mathrm{~J}$
Answer
Correct option: B.
$20 \mathrm{~J}$
$\Delta U=\Delta Q-W=35-15=20 \mathrm{~J}$
View full question & answer→
MCQ 2271 Mark
If a cylinder containing a gas at high pressure explodes, the gas undergoes
  • A
    Reversible adiabatic change and fall of temperature
  • B
    Reversible adiabatic change and rise of temperature
  • ✓
    Irreversible adiabatic change and fall of temperature
  • D
    Irreversible adiabatic change and rise of temperature
Answer
Correct option: C.
Irreversible adiabatic change and fall of temperature
Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature.
View full question & answer→
MCQ 2281 Mark
Heat given to a system is $35$ joules and work done by the system is $15$ joules. The change in the internal energy of the system will be
  • A
    $-50 \mathrm{~J}$
  • ✓
    $20 \mathrm{~J}$
  • C
    $30 \mathrm{~J}$
  • D
    $50 \mathrm{~J}$
Answer
Correct option: B.
$20 \mathrm{~J}$
$\Delta U=\Delta Q-W=35-15=20 \mathrm{~J}$
View full question & answer→
MCQ 2291 Mark
If a cylinder containing a gas at high pressure explodes, the gas undergoes
  • A
    Reversible adiabatic change and fall of temperature
  • B
    Reversible adiabatic change and rise of temperature
  • ✓
    Irreversible adiabatic change and fall of temperature
  • D
    Irreversible adiabatic change and rise of temperature
Answer
Correct option: C.
Irreversible adiabatic change and fall of temperature
Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature.
View full question & answer→
MCQ 2301 Mark
Which of the accompanying $P V$, diagrams best represents an isothermal process
  • A
    Image
  • ✓
    Image
  • C
    Image
  • D
    Image
Answer
Correct option: B.
Image
In isothermal process $P \propto \frac{1}{V}$.Hence graph between $P$ and $V$ is a hyperbola.
View full question & answer→
MCQ 2311 Mark
Which of the accompanying $P V$, diagrams best represents an isothermal process
  • A
    Image
  • ✓
    Image
  • C
    Image
  • D
    Image
Answer
Correct option: B.
Image
In isothermal process $P \propto \frac{1}{V}$.Hence graph between $P$ and $V$ is a hyperbola.
View full question & answer→
MCQ 2321 Mark
Which of the accompanying $P V$, diagrams best represents an isothermal process
  • A
    Image
  • ✓
    Image
  • C
    Image
  • D
    Image
Answer
Correct option: B.
Image
In isothermal process $P \propto \frac{1}{V}$.Hence graph between $P$ and $V$ is a hyperbola.
View full question & answer→
MCQ 2331 Mark
$1 \mathrm{~cm}^3$ of water at its boiling point absorbs $540$ calories of heat to become steam with a volume of $1671 \mathrm{~cm}^3$.If the atmospheric pressure $=1.013 \times 10^5 \mathrm{~N} / \mathrm{m}^2$ and the mechanical equivalent of heat $=4.19 \mathrm{~J} /$ calorie, the energy spent in this process in overcoming intermolecular forces is
  • A
    $540 \mathrm{cal}$
  • B
    $40 \mathrm{cal}$
  • ✓
    $500 \mathrm{cal}$
  • D
    Zero
Answer
Correct option: C.
$500 \mathrm{cal}$
$\Delta Q=\Delta U+\Delta W $
$\therefore \Delta U=\Delta Q-\Delta W=540-\frac{P\left(V_2 / V_1\right)}{J} $
$=540-\frac{1.013 \times 10^5 \times\left[(1671-1) \times 10^{-6}\right]}{4.2} $
$=540-39.7$
$=500 \text { calories }$
View full question & answer→
MCQ 2341 Mark
$1 \mathrm{~cm}^3$ of water at its boiling point absorbs $540$ calories of heat to become steam with a volume of $1671 \mathrm{~cm}^3$.If the atmospheric pressure $=1.013 \times 10^5 \mathrm{~N} / \mathrm{m}^2$ and the mechanical equivalent of heat $=4.19 \mathrm{~J} /$ calorie, the energy spent in this process in overcoming intermolecular forces is
  • A
    $540 \mathrm{cal}$
  • B
    $40 \mathrm{cal}$
  • ✓
    $500 \mathrm{cal}$
  • D
    Zero
Answer
Correct option: C.
$500 \mathrm{cal}$
$\Delta Q=\Delta U+\Delta W $
$\therefore \Delta U=\Delta Q-\Delta W=540-\frac{P\left(V_2 / V_1\right)}{J} $
$=540-\frac{1.013 \times 10^5 \times\left[(1671-1) \times 10^{-6}\right]}{4.2} $
$=540-39.7$
$=500 \text { calories }$
View full question & answer→
MCQ 2351 Mark
$1 \mathrm{~cm}^3$ of water at its boiling point absorbs $540$ calories of heat to become steam with a volume of $1671 \mathrm{~cm}^3$.If the atmospheric pressure $=1.013 \times 10^5 \mathrm{~N} / \mathrm{m}^2$ and the mechanical equivalent of heat $=4.19 \mathrm{~J} /$ calorie, the energy spent in this process in overcoming intermolecular forces is
  • A
    $540 \mathrm{cal}$
  • B
    $40 \mathrm{cal}$
  • ✓
    $500 \mathrm{cal}$
  • D
    Zero
Answer
Correct option: C.
$500 \mathrm{cal}$
$\Delta Q=\Delta U+\Delta W $
$\therefore \Delta U=\Delta Q-\Delta W=540-\frac{P\left(V_2 / V_1\right)}{J} $
$=540-\frac{1.013 \times 10^5 \times\left[(1671-1) \times 10^{-6}\right]}{4.2} $
$=540-39.7$
$=500 \text { calories }$
View full question & answer→
MCQ 2361 Mark
In an adiabatic expansion of a gas initial and final temperatures are $T_1$ and $T_2$ respectively, then the change in internal energy of the gas is
  • ✓
    $\frac{R}{\gamma-1}\left(T_2-T_1\right)$
  • B
    $\frac{R}{\gamma-1}\left(T_1-T_2\right)$
  • C
    $R\left(T_1-T_2\right)$
  • D
    Zero
Answer
Correct option: A.
$\frac{R}{\gamma-1}\left(T_2-T_1\right)$
$\Delta U=-\Delta W=-\frac{R(T_1-T_2)}{(\gamma-1)}=\frac{R(T_2T_1)}{\gamma-1}$
View full question & answer→
MCQ 2371 Mark
In an adiabatic expansion of a gas initial and final temperatures are $T_1$ and $T_2$ respectively, then the change in internal energy of the gas is
  • ✓
    $\frac{R}{\gamma-1}\left(T_2-T_1\right)$
  • B
    $\frac{R}{\gamma-1}\left(T_1-T_2\right)$
  • C
    $R\left(T_1-T_2\right)$
  • D
    Zero
Answer
Correct option: A.
$\frac{R}{\gamma-1}\left(T_2-T_1\right)$
$\Delta U=-\Delta W=-\frac{R(T_1-T_2)}{(\gamma-1)}=\frac{R(T_2T_1)}{\gamma-1}$
View full question & answer→
MCQ 2381 Mark
In an adiabatic expansion of a gas initial and final temperatures are $T_1$ and $T_2$ respectively, then the change in internal energy of the gas is
  • ✓
    $\frac{R}{\gamma-1}\left(T_2-T_1\right)$
  • B
    $\frac{R}{\gamma-1}\left(T_1-T_2\right)$
  • C
    $R\left(T_1-T_2\right)$
  • D
    Zero
Answer
Correct option: A.
$\frac{R}{\gamma-1}\left(T_2-T_1\right)$
$\Delta U=-\Delta W=-\frac{R(T_1-T_2)}{(\gamma-1)}=\frac{R(T_2T_1)}{\gamma-1}$
View full question & answer→
MCQ 2391 Mark
During the adiabatic expansion of $2$ moles of a gas, the internal energy was found to have decreased by $100 \mathrm{~J}$. The work done by the gas in this process is
  • A
    Zero
  • B
    $-100 \mathrm{~J}$
  • C
    $200 \mathrm{~J}$
  • ✓
    $100 \mathrm{~J}$
Answer
Correct option: D.
$100 \mathrm{~J}$
For adiabatic process from $\text{FlOT}$
$\Delta W=-\Delta U \quad(\because \Delta Q=0) $
$\Rightarrow \Delta W=-(-100)=+100 J$
View full question & answer→
MCQ 2401 Mark
During the adiabatic expansion of $2$ moles of a gas, the internal energy was found to have decreased by $100 \mathrm{~J}$. The work done by the gas in this process is
  • A
    Zero
  • B
    $-100 \mathrm{~J}$
  • C
    $200 \mathrm{~J}$
  • ✓
    $100 \mathrm{~J}$
Answer
Correct option: D.
$100 \mathrm{~J}$
For adiabatic process from $\text{FlOT}$
$\Delta W=-\Delta U \quad(\because \Delta Q=0) $
$\Rightarrow \Delta W=-(-100)=+100 J$
View full question & answer→
MCQ 2411 Mark
During the adiabatic expansion of $2$ moles of a gas, the internal energy was found to have decreased by $100 \mathrm{~J}$. The work done by the gas in this process is
  • A
    Zero
  • B
    $-100 \mathrm{~J}$
  • C
    $200 \mathrm{~J}$
  • ✓
    $100 \mathrm{~J}$
Answer
Correct option: D.
$100 \mathrm{~J}$
For adiabatic process from $\text{FlOT}$
$\Delta W=-\Delta U \quad(\because \Delta Q=0) $
$\Rightarrow \Delta W=-(-100)=+100 J$
View full question & answer→
MCQ 2421 Mark
When heat is given to a gas in an isothermal change, the result will be
  • ✓
    External work done
  • B
    Rise in temperature
  • C
    Increase in internal energy
  • D
    External work done and also rise in temp.
Answer
Correct option: A.
External work done
In isothermal change, temperature remains constant,
Hence $\Delta U$ $=0$.
Also from $\Delta Q=\Delta U+\Delta W \Rightarrow \Delta Q=\Delta W$
View full question & answer→
MCQ 2431 Mark
When heat is given to a gas in an isothermal change, the result will be
  • ✓
    External work done
  • B
    Rise in temperature
  • C
    Increase in internal energy
  • D
    External work done and also rise in temp.
Answer
Correct option: A.
External work done
In isothermal change, temperature remains constant,
Hence $\Delta U$ $=0$.
Also from $\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta W$
View full question & answer→
MCQ 2441 Mark
When heat is given to a gas in an isothermal change, the result will be
  • ✓
    External work done
  • B
    Rise in temperature
  • C
    Increase in internal energy
  • D
    External work done and also rise in temp.
Answer
Correct option: A.
External work done
In isothermal change, temperature remains constant,
Hence $\Delta U$ $=0$.
Also from $\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q=\Delta W$
View full question & answer→
MCQ 2451 Mark
A polyatomic gas $\left(\gamma=\frac{4}{3}\right)$ is compressed to $\frac{1}{8}$ of its volume adiabatically. If its initial pressure is $P_o$, its new pressure will be
  • A
    $8 P_o$
  • ✓
    $16 P_0$
  • C
    $6 P_0$
  • D
    $2 P_0$
Answer
Correct option: B.
$16 P_0$
$\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma$
$\Rightarrow P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma$
$=P_0(8)^{4 / 3}$
$=16 P_0$.
View full question & answer→
MCQ 2461 Mark
A polyatomic gas $\left(\gamma=\frac{4}{3}\right)$ is compressed to $\frac{1}{8}$ of its volume adiabatically. If its initial pressure is $P_o$, its new pressure will be
  • A
    $8 P_o$
  • ✓
    $16 P_0$
  • C
    $6 P_0$
  • D
    $2 P_0$
Answer
Correct option: B.
$16 P_0$
$\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma$
$\Rightarrow P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma$
$=P_0(8)^{4 / 3}$
$=16 P_0$.
View full question & answer→
MCQ 2471 Mark
A polyatomic gas $\left(\gamma=\frac{4}{3}\right)$ is compressed to $\frac{1}{8}$ of its volume adiabatically. If its initial pressure is $P_o$, its new pressure will be
  • ✓
    $8 P_0$
  • B
    $16 P_0$
  • C
    $6 P_0$
  • D
    $2 P_0$
Answer
Correct option: A.
$8 P_0$
$\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma$
$\Rightarrow P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma$
$=P_0(8)^{4 / 3}$
$=16 P_0$.
View full question & answer→
MCQ 2481 Mark
A system is given $300$ calories of heat and it does $600$ joules of work. How much does the internal energy of the system change in this process$(J=4.18$ joules $/ \mathrm{cal})$
  • ✓
    $654$ Joule
  • B
    $156.5$ Joule
  • C
    $-300$ Joule
  • D
    $-528.2$ Joule
Answer
Correct option: A.
$654$ Joule
$ \Delta Q=\Delta U+\Delta W, $
$\Delta U= \Delta Q-\Delta W $
$\Delta U=4.18 \times 300-600$
$=654 \text { Joule }$
View full question & answer→
MCQ 2491 Mark
A system is given $300$ calories of heat and it does $600$ joules of work. How much does the internal energy of the system change in this process$(J=4.18$ joules $/ \mathrm{cal})$
  • ✓
    $654$ Joule
  • B
    $156.5$ Joule
  • C
    $-300$ Joule
  • D
    $-528.2$ Joule
Answer
Correct option: A.
$654$ Joule
$ \Delta Q=\Delta U+\Delta W, $
$\Delta U= \Delta Q-\Delta W $
$\Delta U=4.18 \times 300-600$
$=654 \text { Joule }$
View full question & answer→
MCQ 2501 Mark
A system is given $300$ calories of heat and it does $600$ joules of work. How much does the internal energy of the system change in this process$(J=4.18$ joules $/ \mathrm{cal})$
  • ✓
    $654$ Joule
  • B
    $156.5$ Joule
  • C
    $-300$ Joule
  • D
    $-528.2$ Joule
Answer
Correct option: A.
$654$ Joule
$ \Delta Q=\Delta U+\Delta W, $
$\Delta U= \Delta Q-\Delta W $
$\Delta U=4.18 \times 300-600$
$=654 \text { Joule }$
View full question & answer→
Generate Paper FreeAll Thermodynamics topics
MCQ - Page 5 - JEE physics STD 12 Science Questions - Vidyadip