Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
Filter By
Clear Filter

All Questions

The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively.  The heat of formation (in kJ) of carbon monoxide per mole is:
Option: 1 110.5
Option: 2 676.5
Option: 3 -676.5
Option: 4 -110
 

\mathrm{C_s+O_2_g \rightarrow CO_2_g \: \: \: \: \Delta H= -393.5 KJmol^{-1}}

CO_{\left ( g \right )}+\frac{1}{2}O_{2} \: _{\left ( g \right )}\rightarrow CO_{2} \:_{(g)} \: \: \: \: \: \Delta H=-283.5 kJmol^{-1}

C_{\left ( s \right )}+\frac{1}{2}O_{2} \:_{\left ( g \right )}\rightarrow CO_{\left ( g \right )}

\therefore \Delta H= -393.5+283.5

            = -110.0\ kJmol^{-1}

Therefore, Option(4) is correct

View Full Answer(1)
Posted by

Ritika Jonwal

A gas undergoes change from state A to state B.  In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively.  Now gas is brought back to A by another process during which 3 J of heat is evolved.  In this reverse process of B to A :
Option: 1  10 J of the work will be done by the gas.
Option: 2  6 J of the work will be done by the gas.
Option: 3 10 J of the work will be done by the surrounding on gas.
Option: 4  6 J of the work will be done by the surrounding on gas.  
 

A \longrightarrow B

\mathrm{Q=5 \ J}

\mathrm{W=8\ J}

\mathrm{\Delta U_{AB}=Q+W=5+(-8)=-3J}

 

B \longrightarrow A

\mathrm{Q=-3J}

\mathrm{\Delta U_{BA}=3J}

\Delta U_{BA}=-3+W

3+3=W

W=6J

(work is done on the system)

Or

As work done has a positive sign, work is done by the surrounding on the gas.

Hence, the correct answer is (4)

View Full Answer(1)
Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

 For the P-V diagram given for an ideal gas, out of the following which one correctly represents the T-P diagram ?  
Option: 1

Option: 2 

Option: 3 

Option: 4 
 

\because PV=constant

\therefore The process is isothermal (temperature constant)

Also, from the process given, it can be seen that pressure is decreasing, P2<P1

Hence, the correct answer is Option (3)

View Full Answer(1)
Posted by

vishal kumar

A steel rail of length 5 m and area of cross section 40 cm2 is prevented from expanding along its length while the temperature rises by 100C.  If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately :
Option: 1 2×107 N
Option: 2  1×105 N
Option: 3  2×109 N
Option: 4  3×10−5 N  
 

\\ \text{Youngs modules} \ \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2} \\ \text{linears expression coefficient} \alpha=1.2 \times 10^{-5} \mathrm{~K}^{-1} \\ 1=5 \mathrm{~cm} \quad \mathrm{~A}=40 \mathrm{~cm}^2 \\ \Delta \mathrm{T}=10^{\circ} \mathrm{C} \\ \Delta \mathrm{l}=\mathrm{l} \propto \Delta \mathrm{T} \\ \frac{\Delta \mathrm{l}}{\mathrm{l}}=\propto \Delta \mathrm{T} \left[\mathrm{F}=\mathrm{YA} \frac{\Delta \mathrm{l}}{\mathrm{l}}\right] \\ \mathrm{F}=\mathrm{YA} \propto \Delta \mathrm{T} \mathrm{F}=2 \times 10^{11} \mathrm{Nm}^{-2} \times \frac{40}{10} \\mathrm{~m} \times 1.25 \times 10^{-5} \times 10 =2 \times 40 \times 12 \times 10^2 =96 \times 10^3 =0.96 \times 10^3 F \cong 1 \times 10^5 \mathrm{~N}

View Full Answer(1)
Posted by

vishal kumar

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

 For a reaction, A(g) → A(\small \l);  \mathrm{\Delta H = -3RT}. The correct statement for the reaction is :
Option: 1 |\Delta \mathrm{H}|<|\Delta \mathrm{U}|
Option: 2 \Delta H= \Delta U= 0
Option: 3 |\Delta \mathrm{H}|>|\Delta \mathrm{U}|

Option: 4 |\Delta \mathrm{H}|=|\Delta \mathrm{U}|

View Full Answer(1)
Posted by

vishal kumar

 In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 1500C. Immediately, it is put into water of volume 150 cc at 270C kept in a calorimeter of water equivalent to 0.025 kg.  Final temperature of the system is 400C.  The specific heat of aluminium is : (take 4.2 Joule=1 calorie)
Option: 1 378 J/kg – 0C
Option: 2  315 J/kg – 0C
Option: 3  476 J/kg – 0C  
Option: 4  434 J/kg – 0C  
 

\\ \text{Heat lost by metal Blodg} = \text{Heat gain by} \mathrm{H}_2 \mathrm{O}+ \text{contain} \\ \begin{aligned} &\Rightarrow \mathrm{m}_1 \mathrm{C}_1 \Delta \mathrm{T}_1=\left(\mathrm{m}_2+\mathrm{w}\right) \mathrm{C}_2 \Delta \mathrm{T}_2 \\ &\Rightarrow 0.20 \times \mathrm{C}_1 \times(150-40)=\left(150 \times 10^{-3}\right) \\ &\mathrm{C}_1=434 \mathrm{y} \mathrm{kg}^{-1} \mathrm{k}^{-1} \end{aligned}

View Full Answer(1)
Posted by

vishal kumar

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

U is equal to :
Option: 1 Adiabatic work
Option: 2  Isothermal work
Option: 3  Isochoric work
Option: 4 Isobaric work
 

Adiabatic Process -

Heat exchange between the system and surroundings is zero. 

So,

\Delta E= q+w 

q= 0

\Delta E=w

 

No change in internal energy = Adiabatic work

Ans(1)

View Full Answer(1)
Posted by

vishal kumar

Given C(graphite)+O2(g) → CO2(g) ; rH0=−393.5 kJ mol−1 H2(g)+ \frac{1}{2} O2(g) → H2O(l) ; rH0=−285.8 kJ mol−1 CO2(g)+2H2O(l) → CH4(g)+2O2(g) ; rH0=+890.3 kJ mol−1 Based on the above thermochemical equations, the value of rH0 at 298 K for the reaction C(graphite)+2H2(g) → CH4(g) will be :
Option: 1  −74.8 kJ mol−1
Option: 2 −144.0 kJ mol−1
Option: 3 +74.8 kJ mol−1
Option: 4 +144.0 kJ mol−1  
 

C+O_{2} \rightarrow CO_{2} , \Delta H = -3.93.5 kJ mol−1

H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O , \Delta H =28.5 kJ mol−1

\Delta 2H_{2}+O_{2}\rightarrow 2H_{2}O, \Delta H= 2\times \left ( -285.8 \right ) kJ mol−1

CO_{2}+2H_{2}O \rightarrow CH_{4}+2O_{2}, \Delta H=890.3 kJ mol−1

Adding all equations, we get

C+2H_{2}\rightarrow CH_{4}

\Delta H = 890.3-2\times 285.8-399.6

= -74.8 kJ mol−1

View Full Answer(1)
Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Among the following, the set of parameters that represents path functions,is :

\left ( A \right )  q+w
\left ( B \right )   q
\left ( C \right )   w
\left ( D \right )  H-TS
Option: 1   \left ( B \right )\: and\: \left ( C \right )
Option: 2   \left ( B \right ),\left ( C \right )and \left ( D \right )
Option: 3  \left ( A \right )\: and \: \left ( D \right )
Option: 4  \left ( A \right ),\left ( B \right ) and \left ( C \right )
 

 Q+w=\Delta U   is state function 

 H-TS=G  is state function

 q and w are path functions

because q and w depend on the path of the reaction. 

Hence, option number (1) is correct.

View Full Answer(1)
Posted by

Ritika Jonwal

A copper ball of mass 100 gm is at a temperature T.  It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 750C.  T is given by : (Given : room temperature=300C, specific heat of copper=0.1 cal/gm0C)    
Option: 1 8000C
Option: 2  8850C  
Option: 3 12500C
Option: 4  8250C
 

As we have learnt

As we know Heat given  = Heat Taken

\\*\Rightarrow 100\times 0.1\times (T-75) = (100\times 0.1\times 45) + (170\times 1\times 45) \\* 10 (T - 75) = 8100 \\* T-75 = 810 \Rightarrow T = 885\degree C

View Full Answer(1)
Posted by

vishal kumar

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE
filter_img