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Heat and Thermodynamics

JEE Physics · 315 questions · Page 24 of 32 · Click an option or "Show Solution" to reveal answer

Q231
The pressure and volume of an ideal gas are related as PV32=K\mathrm{PV}^{\dfrac{3}{2}}=\mathrm{K} (Constant). The work done when the gas is taken from state A(P1,V1,T1)A\left(P_1, V_1, T_1\right) to state B(P2,V2,T2)B\left(P_2, V_2, T_2\right) is :
A 2(P2V2−P1V1)2\left(\mathrm{P}_2 \sqrt{\mathrm{V}_2}-\mathrm{P}_1 \sqrt{\mathrm{V}_1}\right)
B 2(P1 V1−P2 V2)2\left(\sqrt{\mathrm{P}_1} \mathrm{~V}_1-\sqrt{\mathrm{P}_2} \mathrm{~V}_2\right)
C 2(P2 V2−P1 V1)2\left(\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1\right)
D 2(P1 V1−P2 V2)2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right)
Correct Answer
Option D
Solution

To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process.

For a quasi-static process, the work done

WW

is given by the integral of the pressure

PP

with respect to the volume

VV

:

W=∫V1V2PdVW = \int_{V_1}^{V_2} P dV

Given the relationship

PV32=KPV^{\frac{3}{2}} = K

, we can solve for

PP

:

P=KV32P = \frac{K}{V^{\frac{3}{2}}}

Now, substitute

PP

into the work integral and evaluate it:

W=∫V1V2KV32dV=K∫V1V2V−32dVW = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV

To integrate this, we'll use the power rule for integration. The integral of

V−32V^{-\frac{3}{2}}

is

−2V−12-2V^{-\frac{1}{2}}

, so the work done is:

W=K[−2V−12]V1V2=K(−2V2−12+2V1−12)W = K \left[-2V^{-\frac{1}{2}}\right]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} \right)

We can rewrite

V−12V^{-\frac{1}{2}}

as

1V\frac{1}{\sqrt{V}}

:

W=K(−21V2+21V1)W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right)

Since

P2V232=KP_2V_2^{\frac{3}{2}} = K

and

P1V132=KP_1V_1^{\frac{3}{2}} = K

, we can express

KK

in terms of

P1P_1

and

V1V_1

(or

P2P_2

and

V2V_2

, but we’ll use

P1P_1

and

V1V_1

for now):

W=P1V132(−21V2+21V1)W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} \right)
W=−2P1V1V11V2+2P1V1V11V1W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}}
W=−2P1V1V1V2+2P1V1W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1

Since

V1V2=V1V2\frac{\sqrt{V_1}}{\sqrt{V_2}} = \sqrt{\frac{V_1}{V_2}}

, and recalling

P2V232=KP_2V_2^{\frac{3}{2}} = K

once more:

W=−2P1V1V1V2+2P1V1W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1
W=−2P1V132V2+2P1V1W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1

Putting

P1V132P_1V_1^{\frac{3}{2}}

back as

KK

:

W=−2KV2+2P1V1W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1
W=−2P2V2V21V2+2P1V1W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1
W=−2P2V2+2P1V1W = -2P_2V_2 + 2P_1V_1

So the work done is:

W=2P1V1−2P2V2W = 2P_1V_1 - 2P_2V_2

Therefore, the correct answer matching the given options is: Option D:

2(P1 V1−P2 V2)2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right)

Shortcut : For PVx=\mathrm{PV}^{\mathrm{x}}= constant If work done by gas is asked then

W=nRΔT1−x Here x=32∴W=P2 V2−P1 V1−12=2(P1 V1−P2 V2)\begin{aligned} \mathrm{W} & =\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\mathrm{x}} \\\\ & \text{ Here } \mathrm{x}=\frac{3}{2} \\\\ \therefore \mathrm{W} & =\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{-\frac{1}{2}} \\\\ = & 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right) \end{aligned}
Q232
0.08 kg0.08 \mathrm{~kg} air is heated at constant volume through 5∘C5^{\circ} \mathrm{C}. The specific heat of air at constant volume is 0.17 kcal/kg∘C0.17 \mathrm{~kcal} / \mathrm{kg}^{\circ} \mathrm{C} and J=4.18\mathrm{J}=4.18 joule/ cal\mathrm{~cal}. The change in its internal energy is approximately.
A 318 J
B 298 J
C 284 J
D 142 J
Correct Answer
Option C
Solution

To find the change in the internal energy of air when it is heated at constant volume, we use the formula for heat transfer at constant volume, which is given by:

ΔU=mcvΔT\Delta U = m c_v \Delta T

Where:

ΔU\Delta U

is the change in internal energy,

mm

is the mass of the substance (in this case, air),

cvc_v

is the specific heat at constant volume,

ΔT\Delta T

is the change in temperature. Given that: The mass of air,

m=0.08 kg,m = 0.08 \, \text{kg},

The specific heat of air at constant volume,

cv=0.17 kcal/kg∘C,c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C},

The change in temperature,

ΔT=5∘C,\Delta T = 5^{\circ} \text{C},

And the conversion factor from calories to Joules,

1 cal=4.18 J.1 \, \text{cal} = 4.18 \, \text{J}.

First, convert the specific heat from kcal to Joules:

cv=0.17 kcal/kg∘C×1000 cal/kcal×4.18 J/cal=710.6 J/kg∘Cc_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C} \times 1000 \, \text{cal/kcal} \times 4.18 \, \text{J/cal} = 710.6 \, \text{J/kg}^{\circ}\text{C}

Now, substitute the values into the formula:

ΔU=0.08×710.6×5\Delta U = 0.08 \times 710.6 \times 5
ΔU=284 J\Delta U = 284 \, \text{J}

Thus, the change in internal energy is approximately 284 Joules.

Q233
The average kinetic energy of a monatomic molecule is 0.414 eV0.414 \mathrm{~eV} at temperature : (Use KB=1.38×10−23 J/mol−KK_B=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K})
A 3000 K
B 3200 K
C 1600 K
D 1500 K
Correct Answer
Option B
Solution

To find the temperature at which the average kinetic energy of a monatomic molecule is

0.414 eV0.414 \, \text{eV}

, we use the equation for the average kinetic energy of a molecule in terms of temperature:

Kavg=32kBTK_{\text{avg}} = \frac{3}{2} k_B T

Where:

KavgK_{\text{avg}}

is the average kinetic energy

kBk_B

is the Boltzmann constant, given as

1.38×10−23 J/K1.38 \times 10^{-23} \, \text{J/K}
TT

is the temperature in Kelvin First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules.

The conversion factor is

1 eV=1.6×10−19 J1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}

, so:

Kavg=0.414 eV=0.414×1.6×10−19 JK_{\text{avg}} = 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J}

Substituting

KavgK_{\text{avg}}

and

kBk_B

into the equation:

32kBT=0.414×1.6×10−19 J\frac{3}{2} k_B T = 0.414 \times 1.6 \times 10^{-19} \, \text{J}

Solving for

TT

:

T=0.414×1.6×10−19 J32×1.38×10−23 J/KT = \frac{0.414 \times 1.6 \times 10^{-19} \, \text{J}}{\frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K}}

After performing the calculations:

T≈3200 KT \approx 3200 \, \text{K}

Therefore, the temperature at which the average kinetic energy of a monatomic molecule is

0.414 eV0.414 \, \text{eV}

is approximately 3200 K.

Q234
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of CpCv\dfrac{\mathrm{Cp}}{\mathrm{Cv}} for the gas is :
A 75\dfrac{7}{5}
B 32\dfrac{3}{2}
C 97\dfrac{9}{7}
D 53\dfrac{5}{3}
Correct Answer
Option B
Solution

For an adiabatic process, the following relation holds:

PVγ=constantP V^{\gamma} = \text{constant}

where P is the pressure, V is the volume, and

γ=CpCv\gamma = \frac{C_p}{C_v}

. We are given that the pressure is proportional to the cube of the absolute temperature:

P∝T3P \propto T^3

. Using the ideal gas law,

PV=nRTPV = nRT

, we can rewrite this as:

V∝TP∝TT3∝1T2V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}

. Substituting this into the adiabatic relation, we get:

P(1T2)γ=constantP \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}

. Simplifying, we have:

P1−γT2γ=constantP^{1-\gamma} T^{2\gamma} = \text{constant}

. Since P is proportional to

T3T^3

, we can write:

(T3)1−γT2γ=constant(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}

. This simplifies to:

T3−3γ+2γ=constantT^{3-3\gamma + 2\gamma} = \text{constant}

. For this equation to hold, the exponent of T must be zero. Therefore:

3−3γ+2γ=03 - 3\gamma + 2\gamma = 0

. Solving for γ\gamma, we get:

γ=CpCv=32\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}

. Therefore, the correct answer is Option B.

Q235
NN moles of a polyatomic gas (f=6)(f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of NN is :
A 6
B 2
C 4
D 3
Correct Answer
Option C
Solution
feq=n1f1+n2f2n1+n2\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}

For diatomic gas

feq=5\mathrm{f}_{\mathrm{eq}}=5
5=(N)(6)+(2)(3)N+25 N+10=6 N+6 N=4\begin{aligned} & 5=\frac{(\mathrm{N})(6)+(2)(3)}{\mathrm{N}+2} \\ & 5 \mathrm{~N}+10=6 \mathrm{~N}+6 \\ & \mathrm{~N}=4 \end{aligned}
Q236
The equation of state of a real gas is given by (P+aV2)(V−b)=RT\left(\mathrm{P}+\dfrac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}, where P,V\mathrm{P}, \mathrm{V} and T\mathrm{T} are pressure, volume and temperature respectively and R\mathrm{R} is the universal gas constant. The dimensions of ab2\dfrac{\mathrm{a}}{\mathrm{b}^2} is similar to that of :
A P
B RT
C PV
D R
Correct Answer
Option A
Solution
[P]=[aV2]⇒[a]=[PV2] And [V]=[b][a][b2]=[PV2][V2]=[P]\begin{aligned} & {[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^2\right]} \\ & \text{ And }[\mathrm{V}]=[\mathrm{b}] \\ & \frac{[\mathrm{a}]}{\left[\mathrm{b}^2\right]}=\frac{\left[\mathrm{PV}^2\right]}{\left[\mathrm{V}^2\right]}=[\mathrm{P}] \end{aligned}
Q237
The total kinetic energy of 1 mole of oxygen at 27∘C27^{\circ} \mathrm{C} is : [Use universal gas constant (R)=8.31 J/(R)=8.31 \mathrm{~J} / mole K]
A 6232.5 J
B 5670.5 J
C 6845.5 J
D 5942.0 J
Correct Answer
Option A
Solution

The total kinetic energy of a mole of an ideal gas can be determined by using the equipartition theorem, which states that the energy is equally distributed among degrees of freedom.

For a diatomic molecule such as oxygen (

O2O_2

), there are 5 degrees of freedom (3 translational and 2 rotational - assuming the vibrational modes are not excited at room temperature), so each degree of freedom has an average energy of

12kT\frac{1}{2} kT

per molecule, where

kk

is the Boltzmann constant and

TT

is the temperature in kelvins. However, since we are dealing with moles, we'll use the universal gas constant

RR

instead of the Boltzmann constant

kk

, because

R=k⋅NAR = k \cdot N_A

where

NAN_A

is the Avogadro constant (the number of molecules in a mole).

Therefore, the average energy per mole for each degree of freedom is

12RT\frac{1}{2}RT

.

To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:

Etotal=degrees of freedom×12RTE_{\text{total}} = \text{degrees of freedom} \times \frac{1}{2} R T

For diatomic oxygen:

Etotal=5×12RTE_{\text{total}} = 5 \times \frac{1}{2} R T

Given that temperature

TT

is

27∘C27^{\circ} \mathrm{C}

, we first convert it to kelvins:

TK=TC+273.15=27+273=300 KT_{\text{K}} = T_{\text{C}} + 273.15 = 27 + 273 = 300 \text{ K}

Now we plug in the values for

RR

and

TKT_{\text{K}}

:

Etotal=5×12×8.31 J/mol⋅K×300 KE_{\text{total}} = 5 \times \frac{1}{2} \times 8.31 \text{ J/mol} \cdot \text{K} \times 300 \text{ K}

When we calculate this, we find:

Etotal=52×8.31×300E_{\text{total}} = \frac{5}{2} \times 8.31 \times 300
Etotal=6232.5 J/molE_{\text{total}} = 6232.5 \text{ J/mol}

So the total kinetic energy of 1 mole of oxygen at

27∘C27^{\circ} \mathrm{C}

is approximately 6232.5 J.

Q238
The speed of sound in oxygen at S.T.P. will be approximately: (given, R=8.3 JK−1,γ=1.4R=8.3 \mathrm{~JK}^{-1}, \gamma=1.4)
A 341 m/s
B 333 m/s
C 325 m/s
D 315 m/s
Correct Answer
Option D
Solution

The speed of sound in a gas at standard temperature and pressure (STP) can be calculated using the following formula derived from the ideal gas law and the speed of sound relation in a gas:

v=γRTMv = \sqrt{\gamma \frac{R T}{M}}

Where:

vv

= speed of sound in the gas γ\gamma = adiabatic index (ratio of specific heats,

Cp/CvC_p/C_v

)

RR

= universal gas constant

TT

= temperature in Kelvin

MM

= molar mass of the gas in kilograms per mole (kg/mol) Given:

R=8.3 J⋅mol−1⋅K−1R = 8.3 \, J \cdot mol^{-1} \cdot K^{-1}
γ=1.4\gamma = 1.4

(For diatomic gases such as oxygen)

T=273.15 KT = 273.15 \, K

(Standard temperature, 0°C in Kelvin) Molar mass of Oxygen (

O2O_2

)

M=32×10−3 kg/molM = 32 \times 10^{-3} \, kg/mol

(32 g/mol converted to kg/mol) Plugging these values into the formula:

v=1.4×8.3×273.1532×10−3v = \sqrt{1.4 \times \frac{8.3 \times 273.15}{32 \times 10^{-3}}}

Calculating the values inside the square root:

v=1.4×2268.7450.032v = \sqrt{1.4 \times \frac{2268.745}{0.032}}
v=1.4×70896.40625v = \sqrt{1.4 \times 70896.40625}
v=99304.96875v = \sqrt{99304.96875}
v≈315 m/sv \approx 315 \, m/s
Q239
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:
A 29 RT
B 27 RT
C 20 RT
D 21 RT
Correct Answer
Option B
Solution

To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes

12RT\frac{1}{2} RT

to the internal energy per mole, where RR is the gas constant and TT is the temperature.

An argon atom, being a noble gas, is monoatomic, with 3 translational degrees of freedom.

Since we neglect all vibrational modes, and monoatomic gases have no rotational or vibrational degrees of freedom that contribute to energy at our specified temperature, each mole of argon has

3×12RT=32RT3 \times \frac{1}{2} RT = \frac{3}{2} RT

of energy.

Oxygen, on the other hand, is a diatomic molecule.

This implies that under normal conditions, it has 3 translational and 2 rotational degrees of freedom.

So, each mole of oxygen has

5×12RT=52RT5 \times \frac{1}{2} RT = \frac{5}{2} RT

of energy as we are neglecting vibrational modes which typically become relevant only at higher temperatures.

Hence, the total internal energy (UU) of the gas mixture is:

U=(Energy per mole of Ar×Number of moles of Ar)+(Energy per mole of O2×Number of moles of O2)U = (\text{Energy per mole of Ar} \times \text{Number of moles of Ar}) + (\text{Energy per mole of O}_{2} \times \text{Number of moles of O}_{2})
U=(32RT×8)+(52RT×6)U = \left(\frac{3}{2} RT \times 8\right) + \left(\frac{5}{2} RT \times 6\right)
U=(12RT+15RT)U = \left(12 RT + 15 RT\right)
U=27RTU = 27 RT

Thus, the total internal energy of the system is 27RT27 RT.

Q240
Two vessels AA and BB are of the same size and are at same temperature. A contains 1 g1 \mathrm{~g} of hydrogen and BB contains 1 g1 \mathrm{~g} of oxygen. PA\mathrm{P}_{\mathrm{A}} and PB\mathrm{P}_{\mathrm{B}} are the pressures of the gases in A\mathrm{A} and B\mathrm{B} respectively, then PAPB\dfrac{P_A}{P_B} is:
A 4
B 32
C 8
D 16
Correct Answer
Option D
Solution
PAVAPBVB=nARTAnBRTB Given VA=VB And TA=TBPAPB=nAnBPAPB=1/21/32=16\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}} R T_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}} \mathrm{RT}_{\mathrm{B}}} \\ & \text{ Given } \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \\ & \text{ And } \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}} \\ & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{1 / 2}{1 / 32}=16 \end{aligned}
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