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Heat and Thermodynamics

JEE Physics · 315 questions · Page 13 of 32 · Click an option or "Show Solution" to reveal answer

Q121
The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U == 3PV + 4. The gas is :
A either monoatomic or diatomic.
B monoatomic only.
C polyatomic only.
D diatomic only.
Correct Answer
Option C
Solution

U = 3PV + 4 ⇒\Rightarrow

nf2{{nf} \over 2}

RT = 3PV + 4 ⇒\Rightarrow

f2{{f} \over 2}

PV = 3PV + 4 ⇒\Rightarrow f = 6 +

8PV{8 \over {PV}}

Since degree of freedom is more than 6 therefore gas is polyatomic.

Q122
In thermodynamics, heat and work are :
A Path functions
B Point functions
C Extensive thermodynamics state variables
D Intensive thermodynamic state variables
Correct Answer
Option A
Solution

Heat and work are path function. Heat and work depends on the path taken to reach the final state from initial state.

Q123
The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :
A 3RTV{{3RT} \over V}
B 4RTV{{4RT} \over V}
C 88RTV{{88RT} \over V}
D 52RTV{5 \over 2}{{RT} \over V}
Correct Answer
Option D
Solution

No. of moles of O2 : n1 =

1632{{16} \over {32}}

= 0.5 mole No. of moles of N2 : n2 =

2828{{28} \over {28}}

= 1 mole No. of moles of CO2 : n3 =

4444{{44} \over {44}}

= 1 mole Total no. of moles in container : n = n1 + n2 + n3 ∴\therefore n = 0.5 + 1 + 1 =

52{5 \over 2}

moles Now; PV = nRT ⇒\Rightarrow P =

nRTV{{nRT} \over V}

⇒\Rightarrow P =

52RTV{5 \over 2}{{RT} \over V}
Q124
Calculate the value of mean free path (λ\lambda) for oxygen molecules at temperature 27∘^\circC and pressure 1.01 ×\times 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 ×\times 10−-23 JK−-1)
A 32 nm
B 58 nm
C 86 nm
D 102 nm
Correct Answer
Option D
Solution
Imean=RT2πd2NAP{I_{mean}} = {{RT} \over {\sqrt 2 \pi {d^2}{N_A}P}}
=1.38×300×10−232×3.14×(0.3×10−9)2×1.01×105= {{1.38 \times 300 \times {{10}^{ - 23}}} \over {\sqrt 2 \times 3.14 \times {{(0.3 \times {{10}^{ - 9}})}^2} \times 1.01 \times {{10}^5}}}
=102×10−9= 102 \times {10^{ - 9}}

m

=102= 102

nm

Q125
A polyatomic ideal gas has 24 vibrational modes. What is the value of γ\gamma?
A 1.37
B 1.30
C 1.03
D 10.3
Correct Answer
Option C
Solution

f = 3T + 3R + 24V = 30 γ\gamma = 1 +

2f{2 \over f}

γ\gamma = 1 +

230{2 \over 30}

= 1.066 Nearest Ans. = 1.03

Q126
Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2, m1 and m2, n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :
A n1F1T1+n2F2T2F1+F2{{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{F_1} + {F_2}}}
B n1F1T1+n2F2T2n1F1+n2F2{{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1}{F_1} + {n_2}{F_2}}}
C n1T1+n2T2n1+n2{{{n_1}{T_1} + {n_2}{T_2}} \over {{n_1} + {n_2}}}
D n1F1T1+n2F2T2n1+n2{{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \over {{n_1} + {n_2}}}
Correct Answer
Option B
Solution

Initial internal energy = Final internal energy

F12n1RT1+F22n2RT2=F12n1RT+F22n2RT{{{F_1}} \over 2}{n_1}R{T_1} + {{{F_2}} \over 2}{n_2}R{T_2} = {{{F_1}} \over 2}{n_1}RT + {{{F_2}} \over 2}{n_2}RT

⇒\Rightarrow

T=F1n1T1+F2n2T2F1n1+F2n2T = {{{F_1}{n_1}{T_1} + {F_2}{n_2}{T_2}} \over {{F_1}{n_1} + {F_2}{n_2}}}
Q127
Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is :
A 2K1K2K1+K2{{2{K_1}{K_2}} \over {{K_1} + {K_2}}}
B K1+K2K1K2{{{K_1} + {K_2}} \over {{K_1}{K_2}}}
C K1+K22K1K2{{{K_1} + {K_2}} \over {2{K_1}{K_2}}}
D K1K2K1+K2{{{K_1}{K_2}} \over {{K_1} + {K_2}}}
Correct Answer
Option A
Solution
Req=R1+R2{R_{eq}} = {R_1} + {R_2}

⇒\Rightarrow

1Keq2lA=lK1A+lK2A{1 \over {{K_{eq}}}}{{2l} \over A} = {l \over {{K_1}A}} + {l \over {{K_2}A}}

⇒\Rightarrow

2Keq=lK1+lK2{2 \over {{K_{eq}}}} = {l \over {{K_1}}} + {l \over {{K_2}}}

⇒\Rightarrow

2Keq=K1+K2K1K2{2 \over {{K_{eq}}}} = {{{K_1} + {K_2}} \over {{K_1}{K_2}}}

⇒\Rightarrow

Keq=2K1K2K1+K2{K_{eq}} = {{2{K_1}{K_2}} \over {{K_1} + {K_2}}}
Q128
A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is :
A 2400 J
B 1600 J
C 1800 J
D 3200 J
Correct Answer
Option A
Solution
η=1−12\eta = 1 - {1 \over 2}
η=12\eta = {1 \over 2}
WQ=η{W \over Q} = \eta
1200Q=12{{1200} \over Q} = {1 \over 2}
Q=2400JQ = 2400J
Q129
The ratio of specific heats (CPCV)\left( {{{{C_P}} \over {{C_V}}}} \right) in terms of degree of freedom (f) is given by :
A (1+f3)\left( {1 + {f \over 3}} \right)
B (1+2f)\left( {1 + {2 \over f}} \right)
C (1+f2)\left( {1 + {f \over 2}} \right)
D (1+1f)\left( {1 + {1 \over f}} \right)
Correct Answer
Option B
Solution
CPCV=γ{{{C_P}} \over {{C_V}}} = \gamma
CV=(f2)R{C_V} = \left( {{f \over 2}} \right)R

and

CP−CV=R{C_P} - {C_V} = R
⇒CPC=1+f/2f/2=1+2f\Rightarrow {{{C_P}} \over C} = {{1 + f/2} \over {f/2}} = 1 + {2 \over f}
Q130
If one mole of the polyatomic gas is having two vibrational modes and β\beta is the ratio of molar specific heats for polyatomic gas (β=CPCV)\left( {\beta = {{{C_P}} \over {{C_V}}}} \right) then the value of β\beta is :
A 1.02
B 1.35
C 1.2
D 1.25
Correct Answer
Option C
Solution

For polyatomic gas molecule has 3 rotational degrees of freedom, 3 translational degrees of freedom, and 2 vibrational modes.

So, number of vibrational degrees of freedom = 2 ×\times 2 = 4 Degree of freedom of polyatomic gas f = T + R + V f = 3 + 3 + 4 = 10

β=1+2f=1+210\beta = 1 + {2 \over f} = 1 + {2 \over 10}
β=1210=1.2\beta = {{12} \over 10} = 1.2
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