Ask a Doubt
Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

solve
question image
Asked by adjacentcalliber10 | 10 May, 2019, 12:00: PM
answered-by-expert Expert Answer
for adiabatic process, we have     PV γ = K  ...................(1)
 
where P is pressure, V is volume, γ is ratio of specific heat of constant volume and constant pressure and K is a constant.
 
by taking logarithm on both sides of eqn.(1), we have   ln(P) + γ ln(V) = ln(K)
 
Hence  we have,    ln(P) = ln(K) - γ ln(V) ..............................(2)
 
Eqn.(2) shows the linear relationship between ln(P) and ln(V) as shown in graphs of figure given in question with negative slope.
 
Hence slope of each graph equals the ratio of specific heat of constant volume and constant pressure.
 
slope of graph-X that represent gas-1 is greater than slope of graph-Y that represent gas-2.
 
Hence ratio of specific heats of gas-1 is graeter than that of gas-2
 
γ1 > γ2   ;
 
   begin mathsize 12px style C subscript P 1 end subscript over C subscript V 1 end subscript space greater than space fraction numerator begin display style C subscript P 2 end subscript end style over denominator begin display style C subscript V 2 end subscript end style end fraction space space space space space semicolon space space space space space space space fraction numerator C subscript V 1 end subscript space plus space R over denominator C subscript V 1 end subscript end fraction space greater than space fraction numerator begin display style C subscript V 2 end subscript space plus space R end style over denominator begin display style C subscript V 2 end subscript end style end fraction space space space semicolon space space space space 1 space plus space R over C subscript V 1 end subscript space greater than space 1 space plus space fraction numerator begin display style R end style over denominator begin display style C subscript V 2 end subscript end style end fraction space space space space space space fraction numerator begin display style R end style over denominator begin display style C subscript V 1 end subscript end style end fraction space greater than space space fraction numerator begin display style R end style over denominator begin display style C subscript V 2 end subscript end style end fraction space space space semicolon space space space space space C subscript V 2 end subscript space space greater than space C subscript V 1 end subscript end style
 
 
Specific heat is directly prportional to degrees of freedom f .  Since we have CV2 > CV1 ,  then  f2 > f1
Answered by Thiyagarajan K | 12 May, 2019, 05:24: PM

Application Videos

Multiple Choice Questions Practice Test
Answered Unanswered My Questions My Answers
JEE main - Physics
find current
question image
Asked by arivaryakashyap | 23 Apr, 2024, 10:40: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Three equal charges, 2⋅0 × 10 − 6 C each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.
question image
Asked by ratnadeep.dmr003 | 21 Apr, 2024, 11:06: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Time period for a simple harmonic motion (d ^ 2 * x)/(d * t ^ 2) =- mu xis:
Asked by ksahu8511 | 19 Apr, 2024, 11:55: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
find the acceleration
question image
Asked by mohammedimroz | 13 Apr, 2024, 09:48: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
solution of 12th question
question image
Asked by medhamahesh007 | 02 Apr, 2024, 11:11: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
if Bi with a mass number 211 and atomic number 83 undergoes beta decay the resultant daughter nuclei will be
Asked by gundlasumathi93 | 31 Mar, 2024, 02:13: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
four charges are arranged at corners of square abcd as in figur the force on charge kept at centre o is
question image
Asked by chhayasharma9494 | 31 Mar, 2024, 12:47: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
two small conducting spheres each of mass 9×10‐⁴kg qre suspended from the same point by non conducting strings of length 100cm. they are given equal and similar charges until the strings are rqually inclined at 45° each to the vertcal. The charge on each sphere is
Asked by archithateja3 | 30 Mar, 2024, 10:23: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
how to draw straight line graph and how to identify y-intercept and slope
Asked by Machinenineha | 27 Mar, 2024, 05:28: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Help me
question image
Asked by nandichandi33 | 24 Mar, 2024, 09:21: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
My Class Videos Tests ₹ Plans Ask a Doubt
Get Latest Study Material for Academic year 24-25 Click here
×