datavorous/entrance-exam-dataset huggingface dataset:
An ideal cell of emf 10V is connected in circuit shown in figure. Each resistance is 2Ω. The potential difference (in V) across the capacitor when it is fully charged is _____________

Options:

The correct answer is:
8

Solution:
i=1043+2=10×310=3 Amp
i1=2 A & i2=1 A
VAB=1×2+3×2=8 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (02 Sep Shift 2)
An ideal coil of $10 \mathrm{H}$ is connected in series with a resistance of $5 \Omega$ and a battery of $5 \mathrm{~V}$.
$2$ second after the connection is made the current flowing in amperes in the circuit is

Options:
A
$(1-\mathrm{e})$

B
$\mathrm{e}$

C
$\mathrm{e}^{-1}$

D
$\left(1-\mathrm{e}^{-1}\right)$

The correct answer is:
$\left(1-\mathrm{e}^{-1}\right)$

Solution:
$\mathrm{i}=\mathrm{i}_0\left(1-\mathrm{e}^{\frac{\mathrm{Rt}}{\mathrm{L}}}\right)$
$=\left(1-\mathrm{e}^{-1}\right)$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2007
An ideal diatomic gas occupies a volume V1 at a pressure P1. The gas undergoes a process in which the pressure is proportional to the volume. If at the end of the process the R.M.S. speed of the gas molecules has doubled its initial value, then the heat supplied to the gas, in the given process, is

Options:
A 7P1V1
B 8P1V1
C 9P1V1
D 10P1V1

The correct answer is:
9P1V1

Solution:
As P∝V, PV-1=constant

comparing the above equation with PVx=constant, we get x=-1
Also, C=CV-Rx-1=52R-R-1-1=3R
But as RMS speed is doubled therefore temperature becomes four times.
Hence, Q=nCΔT=n×3R×3Ti=9nRTi
=9P1V1

Topic: Physics, Thermodynamics, JEE Main
An ideal diatomic gas undergoes a thermodynamic process as shown in the P-V diagram. The process AB is isochoric while the process BC is isothermal. The total heat given to the gas in the process is nearly (use ln2 = 0.7 )

Options:
A 2.5 P0V0
B 1.4 P0V0
C 3.9 P0V0
D 1.1 P0V0

The correct answer is:
3.9 P0V0

Solution:
QAB=âUAB+WABâWAB=0
âUAB=f2nRâTâf2âPV
âUAB=52âPVâQAB=2.5 P0V0
Process BC
QBC=âUBC+WBCâQBC=0+2P0V0ln2=1.4 P0V0
Qnet=QAB+QBC=3.9 P0V0

Topic: Physics, Thermodynamics, JEE Main
An ideal diode is connected in a circuit with resistance R=50 Î© and V=10 V. If an AC input signal shown in the figure is given to the circuit, then the maximum and minimum value of output voltage (without load) is

Options:
A 10 V,-15 V
B 10 V,-25 V
C 25 V,-25 V
D 25 V,-15 V

The correct answer is:
10 V,-25 V

Solution:
For the positive half cycle, the diode is in reverse bias. Due to this there will be no current in the resistor, hence the output voltage

Voutput=10 V

For the negative half cycle,

the potential difference across output =-25 V

Topic: Physics, Semiconductors, JEE Main
An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4cm and 4.8cm , respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is:

Options:
A 916
B 32
C 34
D 81256

The correct answer is:
916

Solution:
Using equation of continuity
A1V1=A2V2
V1V2=A2A1=4.86.42=916

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (07 Jan Shift 2)
An ideal gas (Î³=1.5) is expanded adiabatically. How many times its initial volume should the gas be expanded to reduce the root mean square velocity of molecules by a factor of 2?

Options:
A 4 times
B 16 times
C 8 times
D 2 times

The correct answer is:
16 times

Solution:
vrms=3RTM or vrmsâT
vrms is to reduce two times, i.e. the temperature of the gas will have to reduce four times or
T'T=14
During the adiabatic process,
Tâ¡Vâ¡Î³-1=Tâ¡'Vâ¡'Î³-1
or Vâ¡'V=TT'1Î³-1
=411.5-1=42=16
â´ V'=16 V

Topic: Physics, Thermodynamics, JEE Main
An ideal gas γ=1.5 undergoes a thermodynamic process in which the temperature and pressure of the gas are related as T-1P2 = constant. The molar heat capacity of the gas during the process is

Options:
A 32R
B 43R
C 52R
D 3R

The correct answer is:
52R

Solution:
T-1P2= constant
⇒ PV-1 = constant

For a polytropic process PVx = constant, the molar heat capacity is

C=R1γ-1+11-x

Substituting the values we get

C=R11.5-1+11--1=52R

Topic: Physics, Thermodynamics, JEE Main
An ideal gas at atmospheric pressure is adiabatically compressed so that its density becomes 32 times of its initial value. If the final pressure of gas is 128 atmospheres, the value of ' $\gamma$ 'of the gas is :

Options:
A
$1.5$

B
$1.4$

C
$1.3$

D
$1.6$

The correct answer is:
$1.4$

Solution:
Volume of the gas $\mathrm{v}=\frac{\mathrm{m}}{\mathrm{d}}$ and using $\mathrm{PV}^\gamma=$ constant
$$
\begin{aligned}
& \frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{V}}{\mathrm{V}^{\prime}}=\left(\frac{\mathrm{d}^{\prime}}{\mathrm{d}}\right)^\gamma \\
& \text { or } 128=(32)^\gamma \\
& \therefore \gamma=\frac{7}{5}=1.4
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2013 (22 Apr Online)
An ideal gas can't be liquefied because

Options:
A Its critical temperature is always above $0^{\circ} \mathrm{C}$
B Its molecules are relatively smaller in size
C It solidifies before becoming a liquid
D Forces operative between its molecules are negligible

The correct answer is:
Forces operative between its molecules are negligible

Solution:
An ideal gas can't be liquefied because molecules of ideal gas have not force of attraction between them

Topic: Chemistry, States of Matter, JEE Main
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross-sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is M0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency
[Assume the system is in space.]

Options:
A 12πA2γP0MV0.
B 12πMV0AγP0.
C 12πAγP0V0M
D 12πV0MP0A2γ.

The correct answer is:
12πA2γP0MV0.

Solution:
For an adiabatic process,
Pvγ=constant
Differentiating with respect to v,
dPdvvγ+Pγvγ-1=0
dP=-γPvdv=-γPvAdx

Thus, the acceleration, a=dFM=γPA2MVdx
Comparing with , a=-ω2dx, we get,
ω=γP0A2MV0=2πf
f=12πγP0A2MV0

Topic: Physics, Oscillations, JEE Main, JEE Main 2013 (07 Apr)
An ideal gas expands in volume from $1 \times 10^{-3} \mathrm{~m}^3$ to $1 \times 10^{-2} \mathrm{~m}^3$ at $300 \mathrm{~K}$ against a constant pressure of $1 \times 10^5 \mathrm{Nm}^{-2}$. The work done is

Options:
A
$-900 \mathrm{~J}$

B
$900 \mathrm{~kJ}$

C
$2780 \mathrm{~kJ}$

D
$-900 \mathrm{~kJ}$

The correct answer is:
$-900 \mathrm{~J}$

Solution:
Work done $=-P(\Delta V)$
$$
=-1 \times 10^5\left[10^{-2}-10^{-3}\right]=-900 \mathrm{~J}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas expands in volume from $1 \times 10^{-3} \mathrm{~m}^3$ to $1 \times 10^{-2} \mathrm{~m}^3$ at $300 \mathrm{~K}$ against a constant pressure of $1 \times 10^5 \mathrm{Nm}^{-2}$. The work done is

Options:
A
$-900 \mathrm{~J}$

B
$900 \mathrm{~kJ}$

C
$2780 \mathrm{~kJ}$

D
$-900 \mathrm{~kJ}$

The correct answer is:
$-900 \mathrm{~J}$

Solution:
Work done $=-P(\Delta V)$
$$
=-1 \times 10^5\left[10^{-2}-10^{-3}\right]=-900 \mathrm{~J}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2004
An ideal gas expands isothermally from a volume V1 to V2 and then compressed to original volume V1 adiabatically. The initial pressure is P1 and final pressure is P3. The total work done is W. Then:

Options:
A P3>P1, W>0
B P3<P1, W<0
C P3>P1, W<0
D P3=P1, W=0

The correct answer is:
P3>P1, W<0

Solution:
From the graph it is clear that P3>P1 .
Since the area under adiabatic process BCED is greater than that of the isothermal process ABCD . Therefore net word done
W=Wi+-WA
∵ WA>Wi⇒W<0

Topic: Physics, Thermodynamics, JEE Main
An ideal gas expands isothermally from volume V1 to V2 and then it is adiabatically compressed back to its original volume V1. The initial and final pressures of the gas are P1 and P3 respectively and the net work done by the gas is W, then

Options:
A P3>P1, W>0
B P3<P1, W<0
C P3>P1, W<0
D P3=P1, W=0

The correct answer is:
P3>P1, W<0

Solution:
The two processes are shown in the following P-V diagram:

For iothermal process:
P1V1 = P2V2
i.e., P 1 = V 2 V 1 P 2
For adiabatic process:
P 3 V 1 Î³ = P 2 V 2 Î³
i.e. P 3 = V 2 V 1 Î³ P 2
As Î³ > 1 , hence P3 > P1
Further, as slope of adiabatic curve is greater than that of isothermal process curve, adiabatic curve will lie above the isothermal curve. That is, area under adiabatic curve > area under isothermal curve
i.e., Negative work > Positive work
i.e., W < 0

Topic: Physics, Thermodynamics, JEE Main
An ideal gas goes from state $A$ to state $B$ via three different processes as indicated in the $p-V$ diagram
If $Q_1, Q_2, Q_3$ indicate the heat absorbed by the gas along the three processes and $\Delta U_1, \Delta U_2, \Delta U_3$ indicate the change in internal energy along the three processes respectively, then

Options:
A $Q_1>Q_2>Q_3$ and $\Delta U_1=\Delta U_2=\Delta U_3$
B $Q_3>Q_2>Q_1$ and $\Delta U_1=\Delta U_2=\Delta U_3$
C $Q_1=Q_2=Q_3$ and $\Delta U_1>\Delta U_2>\Delta U_3$
D $Q_3>Q_2>Q_1$ and $\Delta U_1>\Delta U_2>\Delta U_3$

The correct answer is:
$Q_1>Q_2>Q_3$ and $\Delta U_1=\Delta U_2=\Delta U_3$

Solution:
For all process 1,2 and 3
$$
\begin{array}{ll}
& \Delta U=U_B-U_A \text { is same } \\
\therefore & \Delta U_1=\Delta U_2=\Delta U_3 \\
\text { Now, } & \Delta Q=\Delta U+\Delta W
\end{array}
$$
Now, $\Delta W=$ work done by the gas
$$
\therefore \quad \Delta Q_1>\Delta Q_2>\Delta Q_3
$$

Topic: Physics, Thermodynamics, NEET, NEET 2012 (Mains)
An ideal gas goes through a process $A \rightarrow B \rightarrow C \rightarrow A$ cycle. The process, $A \rightarrow B$ is adiabatic. Calculate the work done in the process $A \rightarrow B$.

Options:
A $p_0 V_0$
B $\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{(1-\gamma)}$
C $p_0 V_0 \ln (2)$
D $\frac{p_0 V_0\left(2^{1 / \gamma}-1\right)}{(\gamma-1)}$

The correct answer is:
$\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{(1-\gamma)}$

Solution:
We know that, adiabatic relation,
$$
p V^\gamma=\text { constant }
$$
For adiabatic process $A \rightarrow B$
$$
\begin{aligned}
2 p_0 V_0^\gamma & =p_0 V_1^\gamma \\
\Rightarrow \quad V_1^\gamma & =2 V_0^\gamma \Rightarrow V_1=2^{1 / \gamma} V_0
\end{aligned}
$$
$\therefore$ Work done in the process $A \rightarrow B$ is
$$
\begin{aligned}
W & =\frac{p_0 V_1-2 p_0 V_0}{1-\gamma} \\
& =\frac{p_0 V_0 2^{1 / \gamma}-2 p_0 V_0}{1-\gamma} \\
W & =\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{1-\gamma}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas goes through a reversible cycle a→b→c→d has the V - T diagram shown below. Process d→a and b→c are adiabatic.
The corresponding P - V diagram for the process is (all figures are schematic and not drawn to scale) :

Options:
A
B
C
D

The correct answer is:

Solution:
Is an adiabatic process
TVγ-1=const⇒VT1γ-1=const
⇒ as T increase V decreases at non-uniform rate
In process a→b P = constant as V∝T
In process c→d P′= constant s V∝T
But since slope of V - T graph ∝1P
since slope of ab < slope of cd
⇒Pab>Pcd
Also in adiabatic process d→a as T is increasing V in decreasing
⇒ P is increasing, so P - V diagram is as below

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2015 (10 Apr Online)
An ideal gas has molar heat capacity $C_V$ at constant volume. The gas undergoes a process where in the temperature changes as $T=T_0\left(1+\alpha V^2\right)$, where, $T$ and $V$ are temperature and volume respectively, $T_0$ and $\alpha$ are positive constants. The molar heat capacity $C$ of the gas is given as $C=C_V+R f(V)$, where, $f(V)$ is a function of volume. The expression for $f(V)$ is

Options:
A $\frac{\alpha V^2}{1+\alpha V^2}$
B $\frac{1+\alpha V^2}{2 \alpha V^2}$
C $\alpha V^2\left(1+\alpha V^2\right)$
D $\frac{1}{2 \alpha V^2\left(1+\alpha V^2\right)}$

The correct answer is:
$\frac{1+\alpha V^2}{2 \alpha V^2}$

Solution:
Given, $T=T_0\left(1+\alpha V^2\right)$
$$
\begin{aligned}
& \Rightarrow \quad \frac{d T}{d V}=\mathrm{T}_0 \cdot 2 \alpha V \\
& \Rightarrow \quad d V=\frac{d \mathrm{~T}}{\mathrm{~T}_0 \cdot 2 \alpha V}
\end{aligned}
$$
From first law of thermodynamics,
$$
\begin{array}{cc}
& d Q=d U+d W \\
\Rightarrow & n C \cdot d T=n C_V \cdot d T+p d V \\
\Rightarrow & n C \cdot d T+n C_V \cdot d T+p\left(\frac{d T}{T_0 \cdot 2 \alpha V}\right) \\
\Rightarrow & C=C_V+\left(\frac{p}{n}\right)\left(\frac{1}{T_0 2 \alpha V}\right)
\end{array}
$$
Now as gas is ideal,
or
$$
\begin{aligned}
& p V=n R T \\
& \frac{p}{n}=\frac{R T}{V}=\frac{R\left[T_0\left(1+\alpha V^2\right)\right]}{V}
\end{aligned}
$$
substituting in Eq. (ii), we get,
$$
C=C_V+R \cdot\left(\frac{1+\alpha V^2}{\partial \alpha V^2}\right)
$$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure Cp and at constant volume CV is:

Options:
A 75
B 6
C 72
D 52

The correct answer is:
75

Solution:
γ=CpCV=1+2f=1+25=75.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2017 (08 Apr Online)
An ideal gas has volume $V_0$ at $27^{\circ} \mathrm{C}$. It is heated at constant pressure so that its volume becomes $2 V_0$. The final temperature is

Options:
A $54^{\circ} \mathrm{C}$
B $32.6^{\circ} \mathrm{C}$
C $327^{\circ} \mathrm{C}$
D $150 \mathrm{~K}$

The correct answer is:
$327^{\circ} \mathrm{C}$

Solution:
$\frac{T_2}{T_1}=\frac{V_2}{V_1}=2 \Rightarrow T_2=2 \times T_1=2 \times 300=600 \mathrm{~K}=327^{\circ} \mathrm{C}$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas heat engine operates in a Carnot's cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times$ $10^4 \mathrm{~J}$ at high temperature. The amount of heat converted into work is ...

Options:
A $4.8 \times 10^4 \mathrm{~J}$
B $3.5 \times 10^4 \mathrm{~J}$
C $1.6 \times 10^4 \mathrm{~J}$
D $1.2 \times 10^4 \mathrm{~J}$

The correct answer is:
$1.2 \times 10^4 \mathrm{~J}$

Solution:
$\begin{aligned} \eta= & 1-\frac{T_2}{T_1}=1-\frac{400}{500}=\frac{1}{5} \boxtimes \eta=\frac{W}{Q} \Rightarrow \frac{1}{5}=\frac{W}{Q} \\ & \Rightarrow W=\frac{Q}{5}=\frac{6}{5} \times 10^4=1.2 \times 10^4 \mathrm{~J}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas heat engine operates in a Carnot's cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^{4} \mathrm{~J}$ at high temperature. The amount of heat converted into work is

Options:
A $1.6 \times 10^{4} \mathrm{~J}$
B $1.2 \times 10^{4} \mathrm{~J}$
C $4.8 \times 10^{4} \mathrm{~J}$
D $3.5 \times 10^{4} \mathrm{~J}$

The correct answer is:
$1.2 \times 10^{4} \mathrm{~J}$

Solution:
Given, $T_{1}=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}$
$$
T_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}
$$
Heat input, $H_{i}=6 \times 10^{4} \mathrm{~J}$
Efficiency of Carnot engine,
$$
\begin{aligned}
\eta &=1-\frac{T_{2}}{T_{1}} \\
&=1-\frac{400}{500}=\frac{1}{5}
\end{aligned}
$$
Also, efficiency,
$$
\begin{aligned}
\eta &=\frac{\text { Work output }}{\text { Heat input }}=\frac{W}{6 \times 10^{4}} \\
\Rightarrow \quad W &=\eta \times 6 \times 10^{4}=\frac{1}{5} \times 6 \times 10^{4} \\
&=1.2 \times 10^{4} \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas heat engine operates in a Carnot's cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^{4}$ J at high temperature. The amount of heat converted into work is

Options:
A $4.8 \times 10^{4} \mathrm{~J}$
B $35 \times 10^{4} \mathrm{~J}$
C $1.6 \times 10^{4} \mathrm{~J}$
D $1.2 \times 10^{4} \mathrm{~J}$

The correct answer is:
$1.2 \times 10^{4} \mathrm{~J}$

Solution:
Given, amount of heat absorbed, $Q=6 \times 10^{4} \mathrm{~J}$
Temperature of reservoir, $T_{1}=227^{\circ} \mathrm{C}=227+273 \mathrm{~K}$
$$
=500 \mathrm{~K}
$$
Temperature of $\operatorname{sink}, T_{2}=127^{\circ} \mathrm{C}$
$$
\begin{aligned}
&=127+273 \mathrm{~K} \\
&=400 \mathrm{~K}
\end{aligned}
$$
Efficiency of Carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$
$$
=1-\frac{400}{500}=\frac{1}{5}
$$
Work done by Carnot engine $(W)$
$$
\begin{aligned}
&=\text { Efficiency }(\eta) \times \text { heat absorbed }(Q) \\
W & {\left[\eta=\frac{W}{Q}\right] } \\
W &=\frac{1}{5} \times 6 \times 10^{4} \mathrm{~J}=1.2 \times 10^{4} \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas heat engine operates in a Carnot's cycle between 227 Â°C and 127 Â°C. It absorbs 6Ã104 J at the higher temperature. The amount of heat converted into work is

Options:
A 1.6Ã104 J
B 1.2Ã104 J
C 4.8Ã104 J
D 3.5Ã104 J

The correct answer is:
1.2Ã104 J

Solution:
Using the relation,
WQ1=Q1-Q2 Q1
WQ1=1-Q2Q1
WQ1=1-T2T1 âµQ1Q2=T1T2
W=Q11-T2T1
â´ W=6Ã1041-(127+273)(227+273)
W=6Ã1041-400500
=6Ã104Ã100500
=1.2Ã104 J

Topic: Physics, Thermodynamics, JEE Main
An ideal gas heat engine operates in Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^4 \mathrm{cal}$ of heat at higher temperature. Amount of heat converted to work is

Options:
A $2.4 \times 10^4 \mathrm{cal}$
B $6 \times 10^4 \mathrm{cal}$
C $1.2 \times 10^4 \mathrm{cal}$
D $4.8 \times 10^4 \mathrm{cal}$

The correct answer is:
$1.2 \times 10^4 \mathrm{cal}$

Solution:
The heat converted to work is the amount of heat that remains after going through sink.
From the relation $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$
Given,
$\begin{aligned} Q_1 & =6 \times 10^4 \mathrm{cal}, \\ T_1 & =227+273=500 \mathrm{~K} \\ T_2 & =127+273=400 \mathrm{~K}\end{aligned}$
$\therefore \frac{Q_2}{6 \times 10^4}=\frac{400}{500}$
$\Rightarrow Q_2=\frac{4}{5} \times 6 \times 10^4=4.8 \times 10^4 \mathrm{cal}$
Now, heat converted to work
$\begin{aligned}& =Q_1-Q_2 \\& =6.0 \times 10^4-4.8 \times 10^4 \\
& =1.2 \times 10^4 \mathrm{cal}\end{aligned}$

Topic: Physics, Thermodynamics, JIPMER, JIPMER 2007
An ideal gas heat engine operates in Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^4 \mathrm{cal}$ of heat at higher temperature. Amount of heat converted to work is

Options:
A $2.4 \times 10^4 \mathrm{cal}$
B $6 \times 10^4 \mathrm{cal}$
C $1.2 \times 10^4 \mathrm{cal}$
D $4.8 \times 10^4 \mathrm{cal}$

The correct answer is:
$1.2 \times 10^4 \mathrm{cal}$

Solution:
$\begin{aligned} \eta & =\frac{T_1-T_2}{T_1}=\frac{W}{Q} \Rightarrow W=\frac{Q\left(T_1-T_2\right)}{T_1} \\ = & \frac{6 \times 10^4[(227+273)-(273+127)]}{(227+273)} \\ & =\frac{6 \times 10^4 \times 100}{500}=1.2 \times 10^4 \mathrm{cal}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
An ideal gas heat engine operates in Carnot cycle between \( 227^{\circ} \mathrm{C} \) and \( 127^{\circ} \mathrm{C} \). It absorbs \( 6 \times 10^{4} \) cal of heat at high temperature. Amount of heat converted to work is:

Options:
A \( 1.2 \times 10^{4} \mathrm{cal} \)
B \( 4.8 \times 10^{4} \mathrm{cal} \)
C \( 6 \times 10^{4} \mathrm{cal} \)
D \( 2.4 \times 10^{4} \mathrm{cal} \)

The correct answer is:
\( 1.2 \times 10^{4} \mathrm{cal} \)

Solution:
The performance of heat engine is expressed by means of “efficiency” η which is defined as the ratio of useful work obtained from the engine to the heat supplied to it.

Efficiency ncycle =WQ1=T2-T1T2=100500=0.2
Amount heat converted into work
= n Q 2
=0.2×6×104 cal
=1.2×104 cal

Topic: Physics, Thermodynamics, JEE Main
An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true ?

(A) the mean free path of the molecules decreases

(B) the mean collision time between the molecules decreases.

(C) the mean free path remains unchanged.

(D) the mean collision time relations unchanged.

Options:
A (B) and C
B (A)and (B)
C (C) and (D)
D A and (D)

The correct answer is:
(B) and C

Solution:
As we know mean free path

λ=12NVπd2

N = no. of molecule

V = volume of container

d = diameter of molecule

Velocity constant and no. of molecules are same. So mean free path remains same. As temperature increases no. of collision increases so relaxation time decrease.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (02 Sep Shift 2)
An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is \(\mathrm{S}_1\) and that of the other part is \(\mathrm{S}_2\). Given that \(\mathrm{S}_1>\) \(\mathrm{S}_2\). If the piston is removed then the total entropy of the system will be:

Options:
A \(\mathrm{S}_1 \times \mathrm{S}_2\)
B \(\mathrm{S}_1-\mathrm{S}_2\)
C \(\mathrm{S}_1+\mathrm{S}_2\)
D More than one of the above

The correct answer is:
\(\mathrm{S}_1+\mathrm{S}_2\)

Solution:
As entropy is an extensive property and extensive properties are additive. So, the entropy of the system when the piston is released will be \(\mathrm{S} 1+\mathrm{S} 2\). So, the correct answer is option (3).

Topic: Chemistry, Thermodynamics (C), JEE Main
An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is S1 and that of the other part is S2. Given that S1>S2. If the piston is removed then the total entropy of the system will be:

Options:
A S1×S2
B S1-S2
C S1 S2
D S1+S2

The correct answer is:
S1+S2

Solution:
; S1>S2

After piston is removed

|
; Stotal=S1>S2

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If $T_i$ is the initial temperature and $T_f$ is the final temperature, which of the following statements is correct?

Options:
A
$\left(T_f\right)_{\text {irev }}>\left(T_f\right)_{\text {rev }}$

B
$T_f>T_i$ for reversible process but $T_f=T_i$ for irreversible process

C
$\left(T_{\mathrm{f}}\right)_{\text {rev }}=\left(T_{\mathrm{f}}\right)$ irrev

D
$T_f=T_i$ for both reversible and irreversible processes

The correct answer is:
$\left(T_f\right)_{\text {irev }}>\left(T_f\right)_{\text {rev }}$

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2006
An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1 bar. The work done in kJ is:

Options:
A +10.0
B -2.0
C -0.9
D -9.0

The correct answer is:
-0.9

Solution:
w-Pextv2-v1
⇒-1 bar 10L-1L
⇒-9L bar ∴1L bar =100 J
⇒-900 J
⇒-0.90 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (12 Apr Shift 1)
An ideal gas is compressed to half of its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?

Options:
A Isobaric
B Isochoric
C Isothermal
D Adiabatic

The correct answer is:
Adiabatic

Solution:
Since area under the curve is maximum for the adiabatic process so work done on the gas will be maximum for the adiabatic process.

Topic: Physics, Thermodynamics, JEE Main