datavorous/entrance-exam-dataset huggingface dataset:
One mole of a monatomic gas is taken from a point A to another point N along with the path ACB. The initial temperature at A is T0. Calculate the heat absorbed by the gas in the process A → C → B

Options:
A 11RT02
B 7RT02
C 5RT02
D 13RT02

The correct answer is:
11RT02

Solution:
If TB be the temperature at B, then by gas law
P A V A T A = P B V B T B
â´ T B = P B V B T B T A = 2 P 0 2 V 0 P 0 V 0 T 0
The change in internal energy from A to B
Î U = n C v Î T = 1 Ã 3 R 2 Ã 4 T 0 - T 0
= 9 RT 0 2
Work done in the process A to C
W AC = P Î V = P 0 2 V 0 - V 0
= P 0 V 0 = RT 0
and W CB = 0
â´ Total work done from A â C â B
W AC + W CB = RT 0 + 0 = RT 0
From the first law of thermodynamics,
Q = Î U + W
= 9 RT 0 2 + RT 0
= 1 1 RT 0 2
Thus heat absorbed by the gas from A â C â B is
1 1 RT 0 2

Topic: Physics, Thermodynamics, JEE Main
One mole of a monoatomic gas γ1=53 is mixed with one mole of a diatomic gas γ2=75. What is the value of γ for the mixture?

Options:

The correct answer is:
1.5

Solution:
For the mixture,
γmix=μ1γ1γ1-1+μ2γ2γ2-1μ1γ1-1+μ2γ2-1=1×5353-1+1×7575-1153-1+175-1
=32
=1.5

Topic: Physics, Thermodynamics, JEE Main
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α24R J mol K-1; then the value of α will be _____ . (Assume that the given diatomic gas has no vibrational mode.)

Options:

The correct answer is:
3

Solution:
Molecular specific heat at constant volume in terms of degrees of freedom is given by CV=f2R

Molecular specific heat of mixture at constant volume is CV/mix=n1CV1+n2CV2n1+n2

=1×3R2+3×5R21+3

=9R4=α24R

⇒α=3

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (29 Jul Shift 1)
One mole of a monoatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points $\left(V_{0}, T_{0}\right)$ and $\left(2 V_{0}, 3 T_{0}\right)$ in a ${V} \cdot T$ diagram. What is the value of the heat capacity of the gas at the point $\left(V_{0}, T_{0}\right) ?$

Options:
A $R$
B $\frac{3}{2} R$
C $2 R$
D 0

The correct answer is:
$2 R$

Solution:
Heat capacity of an ideal gas in a thermodynamic process,
$\because$ Ideal gas is monoatomic,
$$
\begin{aligned}
C_{V} &=\frac{f R}{2}=\frac{3 R}{2} \\
C_{\text {process }} &=C_{V}+\frac{p}{n} \cdot \frac{d V}{d T} \\
&=\frac{3}{2} R+\frac{p}{n} \cdot \frac{V_{0}}{2 T} \\
\therefore C_{\text {process }}=\frac{3}{2} R+\frac{n R T}{n 2 T} &[\because p V=n R T] \\
=& \frac{3}{2} R+\frac{R}{2}=2 R
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
One mole of a monoatomic ideal gas undergoes the process A â¶ B, as in the given P-V diagram. The specific heat for this process is

Options:
A 3R2
B 15R7
C 30R7
D 20R7

The correct answer is:
15R7

Solution:
ΔQ=ΔU+W (From 1st law)
Here, ΔU=nCVΔT
Temperature at A,
TA=2P0V0R
At B, TB=16P0V0R
So, ΔU=32R14P0V0R=21P0V0
W=2P03V0+123V02P0 [Area under P-V diagram]
=9P0V0
So, ΔQ=21P0V0+9P0V0
=30P0V0
ΔQ=nCΔT
C=ΔQnΔT=30P0V014P0V0R=157R

Topic: Physics, Thermodynamics, JEE Main
One mole of a monoatomic real gas satisfies the equation pV−b=RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

Options:
A
B
C
D

The correct answer is:

Solution:
Here the given equation is P(V-b)=RT.
On comparing with van der Waals' equation
P+aV2[V-b]=RT
That is a=0.
It means only repulsive forces are present which are contributive only at very close distance so the potential energy will increase abruptly. i.e., graph C is correct.

Topic: Chemistry, States of Matter, JEE Main
One mole of a non-ideal gas undergoes a change of state from (2.0 atm, 3.0 L, 100 K) to (4.0 atm, 5.0 L, 250 K) with a change in internal energy, ΔU = 30.0 Latm . The change on enthalpy of process (in L-atm) is

Options:

The correct answer is:
44

Solution:
ÎH=ÎU+P2â¡V2â¡-P1â¡V1
=30+ 4×5 − 2×3
=44âLatm

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole of a perfect gas in a cylinder fitted with a piston has a pressure $P$, volume $V$ and temperature $T$. If the temperature is increased by $1 \mathrm{~K}$ keeping pressure constant, the increase in volume is

Options:
A $\frac{2 V}{273}$
B $\frac{V}{91}$
C $\frac{V}{273}$
D V

The correct answer is:
$\frac{V}{273}$

Solution:
For isobaric process $\frac{V_2}{V_1}=\frac{T_2}{T_1} \Rightarrow V_2=V \times \frac{274}{273}$
Increase $=\frac{274 V}{273}-V=\frac{V}{273}$

Topic: Physics, Thermodynamics, JEE Main
One mole of a perfect gas, initially at a pressure and temperature of 105 N m-2 and 300 K, respectively, expands isothermally until its volume is doubled and then adiabatically until its volume is again doubled. Find the final pressure of the gas. Given Î³ = 1.4 .

Options:
A

0.52.4Ã105 N m-2

B

0.51.4×105 N m-2

C

0.52.4×106 N m-2

D

0.51.4×106 N m-2

The correct answer is:

0.52.4Ã105 N m-2

Solution:
Let P1, V1, T1 be the initial pressure, volume and the temperature of the gas, respectively. For isothermal expansion (1 to 2), we have
V2 = 2V1 ; P1V1 = P2 V2
P2 = P1V1V2 = P12 = 0.5 × 105 N m-2

T2 = T1 = 300 K
For adiabatic expansion (2 to 3),
P2V2γ=P3V3γ
⇒ V3=2V2
⇒ P3=P2V2V3γ=0.5×105×0.51.4

⇒ P3=0.52.4×105 N m-2

Topic: Physics, Thermodynamics, JEE Main
One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of $44 \mathrm{u}$. The alkene is

Options:
A
propene

B
1-butene

C
2-butene

D
ethene

The correct answer is:
2-butene

Solution:
2-butene is symmetrical alkene
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \frac{\mathrm{O}_3}{\mathrm{Zn} / \mathrm{H}_2 \mathrm{O}} 2 \cdot 2 \cdot \mathrm{CH}_3 \mathrm{CHO}$
Molar mass of $\mathrm{CH}_3 \mathrm{CHO}$ is $44 \mathrm{u}$.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2010
One mole of a van der Waals' gas obeying the equation $\left(p+\frac{a}{V^{2}}\right)(V-b)=R T$ undergoes the quasi-static cyclic process which is shown in the $p \cdot V$ diagram. The net heat absorbed by the gas in this process is

Options:
A $\frac{1}{2}\left(p_{1}-\rho_{2}\right)\left(V_{1}-V_{2}\right)$
B $\frac{1}{2}\left(p_{1}+p_{2}\right)\left(V_{1}-V_{2}\right)$
C $\frac{1}{2}\left(A_{1}+\frac{a}{V_{1}^{2}}-p_{2}-\frac{a}{V_{2}^{2}}\right)\left(V_{1}-V_{2}\right)$
D $\frac{1}{2}\left(\rho_{1}+\frac{a}{V_{1}^{2}}+p_{2}+\frac{a}{V_{2}^{2}}\right)\left(V_{1}-V_{2}\right)$

The correct answer is:
$\frac{1}{2}\left(p_{1}-\rho_{2}\right)\left(V_{1}-V_{2}\right)$

Solution:
For the cyclic process

Heat absorbed = Work done
$$
\begin{array}{l}
=\text { Area }=\frac{1}{2}(\Delta p) \times \Delta V \\
=\frac{1}{2}\left(p_{1}-p_{2}\right) \times\left(V_{1}-V_{2}\right) \\
=\frac{1}{2}\left(p_{1}-p_{2}\right)\left(V_{1}-V_{2}\right)
\end{array}
$$

Topic: Physics, Thermodynamics, JEE Main
One mole of alkene $\underline{X}$ on ozonolysis gave one mole of acetaldehyde and one mole of acetone. The IUPAC name of $\underline{X}$ is

Options:
A 2-methyl-2-butene
B 2-methyl-1-butene
C 2-butene
D 1-butene

The correct answer is:
2-methyl-2-butene

Solution:
To decide the structure of alkene that undergoes ozonolysis, bring the products together in such a way that $\mathrm{O}$ atoms are face to face, and replace $\mathrm{O}$ by double $(=)$ bond. Thus,

Topic: Chemistry, Hydrocarbons, JEE Main
One mole of an ideal diatomic gas is taken through a process whose P-V diagram is shown in the figure. The work done by the gas is

Options:
A πP0V0+2P0V0
B πP0V02+P0V0
C 2πP0V0+P0V0
D πP0V02+P0V0

The correct answer is:
πP0V02+P0V0

Solution:
Work done = Area under the curve
=P0V0+π42P0V0

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal gas as taken from state A to state B by three different processes (a) ACB, (b) ADB and (c) AEB as shown in the P-V diagram. The heat absorbed by the gas is

Options:
A Greater in process (b) than in (a)
B The least in process (b)
C The same in (a) and (c)
D Less in (c) than in (b)

The correct answer is:
Less in (c) than in (b)

Solution:
Heat absorbed by gas in three processes is given by
QACB=ΔU+WACB
QADB=ΔU
QAEB=ΔU+WAEB
The change in internal energy in all the three cases is the same and WACB is positive, WAEB is negative.
Hence QACB>QADB>QAEB

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (ΔSsurr) in JK-1 is (1 L atm = 101.3 J)

Options:
A 5.763
B 1.013
C - 1.013
D - 5.763

The correct answer is:
- 1.013

Solution:
ΔE=q+w
0=q-PextΔV
q=PextΔV=3 atm (2 - 1) L = 3 atm L
=(3×101.3) Joule
ΔSsurr=-qT=3×101.3300=-1.013 Joule/K

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from $1 \mathrm{~L}$ to $10 \mathrm{~L}$ volume. $\Delta \mathrm{U}$ for this process is
(Use $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

Options:
A $1260 \mathrm{~J}$
B $2520 \mathrm{~J}$
C $5040 \mathrm{~J}$
D $0 \mathrm{~J}$

The correct answer is:
$0 \mathrm{~J}$

Solution:
In an isothermal expansion or compression process, the temperature of system remains constant throughout the process. Since, for an ideal gas, $\mathrm{U}$ depends only on temperature, it follows that the value of \(\Delta U\) is zero, during isothermal expansion of gas.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2022 (Phase 2)
One mole of an ideal gas at 350 K is in a 2.0 L vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0 L to 3.0 L against a constant pressure of 4 atm. The change in entropy of the surroundings (∆S) is _____ J K–1 (Nearest integer)

Given: R=8.314 J K-1 mol-1.

Options:

The correct answer is:
3

Solution:
The first law of thermodynamics for isothermal process Q=-W

The W for isothermal reversible process is

W=-nRTlnV2V1

∆S=QT

∆Ssurrounding=-nRlnV2V1

=-1×8.314×2.303×log32

=-3.37 JK-1

≃-3 JK-1

The given process is mentioned as isothermal reversible process hence the ∆Ssurrounding is calculated.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (12 Apr Shift 1)
One mole of an ideal gas at an initial temperature of $T K$ does $6 R$ joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5 / 3$, the final temperature of gas will be

Options:
A $(T+2.4) K$
B $(T-2.4) K$
C $(T+4) K$
D $(T-4) K$

The correct answer is:
$(T-4) K$

Solution:
$W=\frac{R\left(T_i-T_f\right)}{\gamma-1} \Rightarrow 6 R=\frac{R\left(T-T_f\right)}{\left(\frac{5}{3}-1\right)} \Rightarrow T_f=(T-4) K$

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal gas CV=20JK-1mol-1 initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be-

Options:
A W=0;q=5.46kJ
B W=0;q=0
C W=-5.46kJ;q=5.46kJ
D W=5.46kJ;q=5.46kJ

The correct answer is:
W=0;q=5.46kJ

Solution:
T1=273K
T2=2Ã273K
⇒âT=273
Volume is constant
So, âV=0
W=PÎV=PÃ0=0
q=nCVÎT=20Ã273=5460J=5.46kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole of an ideal gas is allowed to expand freely and adiabatically into vacuum until its volume has doubled. A statement which is not true concerning this expression is

Options:
A $\Delta H=0$
B $\Delta S=0$
C $\Delta E=0$
D $W=0$

The correct answer is:
$W=0$

Solution:
$W=0$ is not true.

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. If the work done during the process is $3 \mathrm{~kJ}$, then final temperature of the gas is $\left(C_V=20 \mathrm{~J} / \mathrm{K}\right)$

Options:
A $100 \mathrm{~K}$
B $150 \mathrm{~K}$
C $195 \mathrm{~K}$
D $255 \mathrm{~K}$

The correct answer is:
$150 \mathrm{~K}$

Solution:
We know that work done, $W=C_v\left(T_1-T_2\right)$
$3 \times 1000=20\left(300-T_2\right) ; \therefore 3000=6000-20 T_2$
$\therefore T_2=\frac{3000}{20}=150 \mathrm{~K}$.

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. $\Delta S$ for the process in $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ is
$[\ln 2=0.693$ and $R=8.314, \mathrm{~J} /(\mathrm{mol} \mathrm{K})]$

Options:
A
$6.76$

B
$5.76$

C
$10.76$

D
$8.03$

The correct answer is:
$5.76$

Solution:
For isothermal process $(\Delta T=0)$
$$
\begin{aligned}
\Delta S & =R \ell \mathrm{n} \frac{P_1}{P_2}=8.314 \ell \mathrm{n} 2 \\
& =8.314 \times 0.693=5.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2012 (26 May Online)
One mole of an ideal gas is taken along the process in which PVx = constant. The graph shown represents the variation of molar heat capacity of such a gas with respect to x. The values of c' and x'. respectively, are given by

Options:
A 5 2 R , 5 2
B 5 2 R , 5 3
C 7 2 R , 7 2
D 5 2 R , 7 5

The correct answer is:
5 2 R , 5 3

Solution:
For P V x =constant
C= C V + R −x+1
At x = â , C = 3 2 R
from PV x = constnat
â P 1 / x V = another constant
so at x = â , C v = 3 2 R
hence C = C v = 3 2 R
and then C p = C v + R = 5 2 R
at x = 0, P = constant and C = C'
hence C â² = C p = 5 2 R
at x = x', C = 0, so the process is adiabatic, hence
x â² = C p C v = 5 3

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal gas passes through a process where pressure and volume obey the relation P=Po1-12VoV2 . Here Po and Vo are constants. Calculate the change in the temperature of the gas if its volume changes from Vo to 2Vo .

Options:
A 14P0V0R
B 54P0V0R
C 12P0V0R
D 34P0V0R

The correct answer is:
54P0V0R

Solution:
For ideal gas, PV=nRT
Therefore, nRTV=P01-12V0V2
⇒ T=P0VR1-12V0V2
Therefore, the change in temperature is:
ΔT=Tf-Ti
=P02V0R1-12V02V02-P0VR1-V022V02
=5P0V04R

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (10 Apr Shift 2)
One mole of an ideal monatomic gas at temperature T0 expands slowly according to the law P=kV (k is constant). If the final temperature is 4T0 then heat supplied to gas is

Options:
A 2 RT0
B 32 RT0
C 6 RT0
D RT02

The correct answer is:
6 RT0

Solution:
P=kV
∴ PV-1=k
∴ x=-1
C=Cv+R1-x=3R2+R1+1=2R
Q=nC∆T=2R×3T0=6RT0

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27oC . The magnitude of work done on the gas will be:

Options:
A 300R
B 300Rln2
C 300ln6
D 300Rln7

The correct answer is:
300Rln2

Solution:
Work done on gas=nRTlnPfPi=R(300)ln(2)=300Rln2.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (15 Apr)
One mole of an ideal monatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by:

Options:
A 254 P0V0R
B 58 P0V0R
C 258 P0V0R
D 2516 P0V0R

The correct answer is:
258 P0V0R

Solution:
Equation of the line BC, P=P0-2P0V0(V-2V0), the temperature = P0V- 2P0V2V0+4P0V1 ×R⇒T=P0R5V-2V2V0⇒dTdV=0
⇒5-4VV0=0⇒V=54V0
⇒T=P0R5×5V04-2V0×2516V02⇒T=258P0V0R.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (16 Apr Online)
One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. If the temperature of the gas at state 1 is 300 K and at state 4 is 500 K, then heat exchanged during process 2â3, is

Options:
A 1000R
B 600R
C 750R
D 800R

The correct answer is:
1000R

Solution:
Q23=nCPT3–T2

The processes 1→2 and 3→4 are isochoric in nature. Hence, for these processes T∝P. As a result, we obtain,
T3=1000 K
T2=600 K
Q23=1×5R2×400=1000R

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal monatomic gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100  K and the universal gas constant R=8.0 J mol-1 K-1, then how much is the decrease in its internal energy (in J) ?

Options:

The correct answer is:
900

Solution:
Vi=V
Vf=8V
For adiabatic process {γ=5/3 for monoatomic gas}
T 1 ⋅ V 1 γ−1 = T 2 ⋅ V 2 γ−1
100V23=T28V23
T2=25 K
Loss in internal energy ∆U=nCv∆T=1fR2100-25=12×75=900 J

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, $27^{\circ} \mathrm{C}$. The work done on the gas will be:

Options:
A
$300 R \ln 6$

B
$300 \mathrm{R}$

C
$300 \mathrm{R} \ln 7$

D
$300 \mathrm{R} \ln 2$

The correct answer is:
$300 \mathrm{R} \ln 2$

Solution:
Work done on gas $=\mathrm{nRT} \ell n\left(\frac{p_f}{p_1}\right)$
$$
=\mathrm{R}(300) \ell \mathrm{n}(2)\left(\because \frac{P_f}{p_i}=2 \text { given }\right)
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} \mathrm{C}$ to $100^\circ C$. Then the change in the internal enery of the gas is (Given, $R=8.32 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ )

Options:
A $0.83 \times 10^{3} \mathrm{J}$
B $46 \times 10^{3} \mathrm{J}$
C $2.08 \times 10^{3} \mathrm{J}$
D $1.25 \times 10^{3} \mathrm{J}$

The correct answer is:
$1.25 \times 10^{3} \mathrm{J}$

Solution:
We know that $\Delta U=n C_{v} \Delta T$
$$
\begin{array}{l}
\text { where } \begin{aligned}
\Delta T &=\left(T_{2}-T_{1}\right) \\
T_{1} &=0^{\circ} \mathrm{C}=273 \mathrm{K} \\
T_{2} &=100^{\circ} \mathrm{C}=373 \mathrm{K} \\
n &=1 \text { (monoatomic gas) }
\end{aligned} \\
\begin{aligned}
R=8.32 \mathrm{J} \mathrm{mol}^{-} \mathrm{k}^{-1}=1 \times \frac{3}{2} R \times(373-273) \\
=1 \times \frac{3}{2} \times 8.32 \times 100=3 \times 8.32 \times 50
\end{aligned}
\end{array}
$$
$\Rightarrow 1248 \mathrm{kJ}$
of
$$
=1.25 \times 10^{3} \mathrm{J}
$$

Topic: Physics, Thermodynamics, JEE Main
One mole of an ideal monoatomic gas is mixed with one mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume is (in Calories)

Options:

The correct answer is:
4.00

Solution:
C v,m  =  3 2 R (for monoatomic gas)
C v,m  =  5 2 R (for diatomic gas)
for the mixture, (C v,m ) mix = 3 2 R×1+ 5 2 R×1 1+1
=2R
∴ (C v,m ) mix =4 cal

Topic: Chemistry, States of Matter, JEE Main
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _______ J (nearest integer)

Given : log 2 = 0.3, ln 10 = 2.3

Options:

The correct answer is:
620

Solution:
1 to 2 , Isobaric process
2 to 3, Isochoric process
3 to 1, Isothermal process
Using work under isobaric, isochoric and isothermal processes respectively:
W = W1→2 + W2→3 + W3→1W = ((-P(V2 - V1) + 0 + (-P1V1(lnV2V1))=((-1(40 - 20) + 0 + (-1 × 20 (ln2040))=620 J

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _______ J (nearest integer)

Given : log 2 = 0.3, ln 10 = 2.3

Options:

The correct answer is:
620

Solution:
1 to 2 , Isobaric process
2 to 3, Isochoric process
3 to 1, Isothermal process
Using work under isobaric, isochoric and isothermal processes respectively:
W = W1→2 + W2→3 + W3→1W = ((-P(V2 - V1) + 0 + (-P1V1(lnV2V1))=((-1(40 - 20) + 0 + (-1 × 20 (ln2040))=620 J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (24 Jan Shift 2)
One mole of complex compound CoNH33Cl3 gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl(s). The structure of the complex is

Options:
A CoNH33Cl32NH3
B CoNH34Cl2Cl.NH3
C CoNH34ClCl2.NH3
D CoNH35ClCl2

The correct answer is:
CoNH35ClCl2

Solution:
CoNH35Cl3 is correctly written as CoNH35ClCl2 which on ionization gives 1 + 2 = 3 moles of ions out of which 2 moles of Cl- ions are present which react with 2 moles of AgNO3 to give 2 moles of AgCl(s).

Topic: Chemistry, Coordination Compounds, JEE Main
One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement :

Options:
A The change in internal energy in whole cyclic process is 250 R
B The change in internal energy in the process CA is 700 R.
C The change in internal energy in the process AB is - 350 R
D The change in internal energy in the process BC is - 500 R

The correct answer is:
The change in internal energy in the process BC is - 500 R

Solution:
Δ U CA = nc v T A - T C = 1 × 5 2 R - 2 0 0 = - 5 0 0 R
Δ U AB = nc v T B - T A = 1 × 5 2 R × 4 0 0 = + 1 0 0 0 R
Δ U BC = nc v T C - T B = 1 × 5 2 R - 2 0 0 = - 5 0 0 R

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2014 (06 Apr)
One mole of fluorine is reacted with two moles of hot concentrated $\mathrm{KOH}$. The products formed are $\mathrm{KF}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{O}_2$. The molar ratio of $\mathrm{KF}, \mathrm{H}_2 \mathrm{O}$ and $\mathrm{O}_2$, respectively is

Options:
A $1: 1: 2$
B $2: 1: 0.5$
C $1: 2: 1$
D $2: 1: 2$

The correct answer is:
$2: 1: 0.5$

Solution:
Stoichiometric equation for the reaction is
$$
\mathrm{F}_2+2 \mathrm{KOH} \longrightarrow 2 \mathrm{KF}+\mathrm{H}_2 \mathrm{O}+\frac{1}{2} \mathrm{O}_2
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
One mole of gas having Î³ = 7 / 5 is mixed with 1 mole of a gas having Î³ = 4 / 3 . What will be the Î³ for the mixture?

Options:
A 1 5 1 1
B 5 1 3
C 5 1 1
D 1 5 1 3

The correct answer is:
1 5 1 1

Solution:
For γ = 7 5 , Cv=Rγ-1=R75-1=5R2
Cp=γRγ-1=7/5R75-1=7R2
For γ=43, Cv=3R, Cp=4R
∴ γ mix = 7 2 + 4 5 2 + 3 = 1 5 1 1

Topic: Physics, Thermodynamics, JEE Main
One mole of ideal gas goes through process P=2V21+V2 then change in temperature of the gas when volume changes from V=1 m3 to 2 m3 is:

Options:
A -45R K
B 115R K
C -52R K
D 2 K

The correct answer is:
115R K

Solution:
P=2V21+V2
P1=1
P2=85
ΔT=T2-T1=P2V2nR-P1V1nR=115R

Topic: Physics, Thermodynamics, JEE Main
One mole of ideal monoatomic gas $(\gamma=5 / 30)$ is mixed with one mole of diatomic gas $(\gamma=7 / 5)$. What is $\gamma$ for the mixture? $\gamma$ denotes the ratio of specific heat at constant pressure, to that at constant volume.

Options:
A
$3 / 2$

B
$23 / 15$

C
$35 / 23$

D
$4 / 3$

The correct answer is:
$3 / 2$

Solution:
$\mathrm{Q}=\mathrm{Q}_1+\mathrm{Q}_2$
$\frac{\mathrm{n}_1+\mathrm{n}_2}{\gamma_{\mathrm{m}}-1}=\frac{\mathrm{n}_1}{\gamma_1-1}+\frac{\mathrm{n}_2}{\gamma_2-1}$
$\gamma_{\mathrm{m}}=\frac{3}{2}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2004
One mole of magnesium nitride on the reaction with an excess of water gives

Options:
A
one mole of ammonia

B
two moles of nitric acid

C
two moles of ammonia

D
one mole of nitric acid

The correct answer is:
two moles of ammonia

Solution:
$\mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{NH}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
One mole of magnesium nitride on the reaction with an excess of water gives

Options:
A
one mole of ammonia

B
two moles of nitric acid

C
two moles of ammonia

D
one mole of nitric acid

The correct answer is:
two moles of ammonia

Solution:
$\mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{NH}_3$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2004
One mole of \( \mathrm{X}_{2} \mathrm{H}_{4} \) loses \( 6 \) moles of electrons to form a new compound \( \mathrm{Z} \). Assuming that all the \( \mathrm{X} \) atoms appear in the new compound, what is the oxidation state of \( \mathrm{X} \) in 'Z'? There is no change in the oxidation state of hydrogen and \( \mathrm{X} \) is more electronegative than hydrogen.

Options:
A \( +1 \)
B \( -3 \)
C \( +3 \)
D \( +5 \)

The correct answer is:
\( +1 \)

Solution:
One mole of X2H4 loses 6 electrons, so, it becomes (X2H4)+6.

Hence, the new compound Z=(X2H4)+6.

The oxidation state of 'X' (assuming its oxidation state is a) can be calculated by:

2(a)+4×1=6

a=+1

The oxidation state of X in 'Z' is +1.

Topic: Chemistry, Redox Reactions, JEE Main