Thermodynamics Test

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Thermodynamics Test - 56

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Weekly Quiz Competition

Question 1
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A car is moving with a speed of $$40 $$ km/hr. If the car engine generated 7 kilowatt power, then the resistive force in the path of the car will be:-

A
360 Netwon

B
630 Newton

C
Zero

D
280 Newton

Solution

Power= Force $$\times$$ speed
$$R= Power/Speed$$
Given, $$P=7Kw=7000W$$
Speed= $$40km/h=11.11m/s$$
Resistance= $$7000/11.11= 630.063 N$$
Question 2
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Distribution of energy in the spectrum of a black body can be correctly represented by ?

A
Wein's law

B
Stefan's law

C
Planck's law

D
Kirchhoff's law

Solution

because plancks law explain the energy of distribution of energy at low as well as high tempreture.
Question 3
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Two moles of an ideal gas, undergo the process shown in figure. $$AB$$ and $$CD$$ are adiabatic processes. The adiabatic exponent of the gas is?

A
$$5/3$$

B
$$7/5$$

C
$$4/3$$

D
$$9/7$$

Solution

$$\textbf{Solution}:$$
$$PV^{\gamma} = constant$$
$$P_0 (2V_0)^{\gamma} = \dfrac{P}{32} (16V_0)^{\gamma}$$
$$P_0 V_0^{\gamma} 2^{\gamma} = PV_0^{\gamma}\dfrac{16^{\gamma}}{32}$$
$$2^{\gamma} = \dfrac{2^{4_{\gamma}}}{2^5}$$
$$2^5 = 2^{3_{\gamma}}$$
$$5 = 3^{\gamma}$$
$$\rho = \dfrac{5}{3}$$ monoatomic gas.

$$\textbf{Hence A is the correct option}$$
Question 4
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one mole of an ideal gas at temperature T1 expends according to the law $$P/V^2 = a(constant)$$. the work done by the gas till the temperature of gas become T2 is:

A
$$\dfrac 1 2 \times R (T_2-T_1)$$

B
$$\dfrac 1 3 \times R (T_2-T_1)$$

C
$$\dfrac 1 4 \times R (T_2-T_1)$$

D
$$\dfrac 1 5 \times R (T_2-T_1)$$

Solution

$$\textbf{Solution}:$$

$$W = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} KV dV$$
$$ (\therefore \dfrac{p}{V} = constant )$$
$$\therefore W = \dfrac {1}{2} k (V_2^2 - V_1^2)$$
PV = RT
But p = KV
$$\therefore KV^2 = RT$$

or $$K (V_2^2 - V_1^2) = R (T_2 - T_1)$$
$$\therefore W = \dfrac{R}{2} (T_2 - T_1)$$

$$\textbf{Hence A is the correct option}$$
Question 5
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Which of the following graphs correctly represents the variation of $$\beta =-\left( \dfrac { dv }{ dp } \right) /V$$ with $$P$$ for an ideal gas at constant temperature?

A

B

C

D

Solution

$$\textbf{Solution}:$$
From PV=RT
dV.P+V.dP=0
or $$\dfrac{dV}{dP}= -\dfrac{V}{P}$$
or $$\beta=\dfrac{1}{P}$$ (Rectangular hyperbola)
$$\beta\times P=1$$.

$$\textbf{Hence B is the correct option}$$
Question 6
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When a gas in enclosed in a closed vessel was heated so as to increase its temperature by $$ 25^o C $$ its pressure was seen to have increased by 10% The initial temperature of the gas was nearly

A
$$ 500^\circ C $$

B
$$ 227^\circ C $$

C
$$ 273^\circ C $$

D
$$ 250^\circ C $$

Solution

Given,

Percentage increase in pressure is $$10$$%

Increase in temperature is $${{25}^{o}}C$$

For constant volume,

$$P\,\,\alpha \,T$$

$$\dfrac{{{P}_{2}}}{{{P}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$$

$$\Rightarrow \dfrac{1.1{{P}_{1}}}{{{P}_{1}}}=\dfrac{\left( {{T}_{1}}+25 \right)}{{{T}_{1}}}$$

$$\Rightarrow {{T}_{1}}=\dfrac{25}{0.1}=250{{\,}^{o}}C$$

Hence, initial temperature is $${{250}^{o}}C$$
Question 7
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when an ideal diatomic gas is heated at constant pressure, its initial energy increased by $$50cal$$ then the work done by the gas is:

A
$$30\ cal$$

B
$$50\ cal$$

C
$$70\ cal$$

D
$$20\ cal$$

Solution

Given: Ideal diatomic gas, $$\triangle U$$= 50 cal
Solution:
For ideal diatomic gas $$C_p = \dfrac{7R}{2}$$ and $$C_v=\dfrac{5R}{2}$$
So, $$\gamma=\dfrac{7}{5}$$
We know that, $$\triangle Q= nC_p {\triangle T}$$
Also $$\triangle U= nC_v{\triangle T}$$
So, $$ \dfrac{\triangle Q}{\triangle U}= \dfrac{nC_p \triangle T}{nC_v\triangle T}=\dfrac{C_p}{C_v}=\gamma$$
$$\therefore \dfrac{\triangle Q}{ 50}=\dfrac{7}{5}$$
$$\Rightarrow \triangle Q= 70 cal$$
From first law of thermodynamics, $$\triangle Q=\triangle U+W$$
Putting the values of $$ \triangle Q= 70 cal $$ and $$\triangle U= 50 cal$$, we get,
W=20 cal
Hence, the correct option is D.
Question 8
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A motor pump is delilvering water at certain rate.In order to increase the rate of delivery by100% ,the power of the motor is to be increased by

A
300%

B
200%

C
400%

D
700%

Solution

$$\text{Solution}:$$
The rate at which the energy is consumed is called power. Hence the expression is given as,
$$P = \dfrac{E}{t} = \dfrac{{\dfrac{1}{2}m{v^2}}}{t}$$
Where, E is the energy, m is the mass, v is the velocity and t is the time.

In order to increase the rate of delivery by 100% the velocity will be increased by 100% from the same pipe at the same time.Therefore the velocity increases like,
$$\left( {100 + 100} \right)v = 200v$$

The new power which is increase by 100% is given as,
$$P = \dfrac{{\dfrac{1}{2}m{{\left( {200v} \right)}^2}}}{t} = \dfrac{{\dfrac{1}{2}m \times 40000{v^2}}}{t} = 400\dfrac{{\dfrac{1}{2}m \times 100{v^2}}}{t} = 400P \times 100$$

Hence to increase the rate of delivery by 100%, the power of the motor will be increased by 400%.It is clear that the high power motors will have high velocity in delivering water. So in order to increase the power of the motor, increase the velocity of water.

$$\textbf{Hence C is the correct option}$$
Question 9
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Which of the following statement about a thermodynamic process is wrong?

A
For an adjabatic process $$\triangle E_{int} = -W$$

B
For a constant volume process $$\triangle E_{int} = +Q$$

C
For a cyclic process $$\triangle E_{int} = 0$$

D
For free expansion of a gas $$\triangle E_{int} > 0$$

Solution

Solution:
In option A,For adiabatic process, $$\triangle Q= 0$$,
$$\therefore \triangle E_{int}= -W$$
In option B, for constant volume process $$W=0$$
$$\therefore \triangle E_{int}= Q$$
In option C, for constant volume process $$E_{int}=0$$
In option D, for free expansion as the gas is expanding freely so change in volume is zero. So, $$W=0$$
$$\therefore \triangle E_{int}= Q$$
Hence,correct option is D.
Question 10
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Heat of $$30\ kcal$$ is supplied to a system and $$4200\ J$$ of external work is done on the system so that its volume decreases at constant pressure. What is the change in its internal energy. $$(1 cal = 4.2 J)$$

A
$$1.68 \times 10^5\ J$$

B
$$1.302 \times 10^5\ J$$

C
$$3.302 \times 10^5\ J$$

D
$$4.302 \times 10^5\ J$$

Solution