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In figure, a resistance coil, wired to an external battery is placed inside a thermally insulated cylinder find with a frictionless piston and containing an ideal diatomic gas A current = 200 ma exists in the coil which has a resistance R=350Ω If the pressure of the gas remains constant what should be the speed v (in cm/s of the piston of mass m=10kg)?
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Asked by samarthghogare | 06 May, 2023, 11:17: AM
answered-by-expert Expert Answer
Let us assume the dissipated heat in the resistance coil is absorbed by the enclosed gas in the cylinder.
 
Rate of heat dissipation is
 
 begin mathsize 14px style fraction numerator d Q over denominator d t end fraction space equals space I to the power of 2 to the power of space end exponent R subscript c space equals space 0.2 space cross times space 0.2 space cross times 350 space equals space 14 space W end style ...................... (1)
 
where i is current through coil and Rc is resistance of coil .
 
For diatomic gas that is hetaed at constant pressure , heat dissipation is
 
begin mathsize 14px style fraction numerator d Q over denominator d t end fraction space equals space n space C subscript p subscript space end subscript fraction numerator d T over denominator d t end fraction end style  .............................. (2)
 
where n is number of moles , Cp =(7/2) R  is constant volume specific heat, R is universal gas constant  and
(dT/dt) is rate of change of increase of temperature.
 
From eqn.(1) and eqn.(2), we get
 
begin mathsize 14px style fraction numerator d T over denominator d t end fraction space equals space fraction numerator 14 over denominator n space C subscript p end fraction space equals space fraction numerator 14 over denominator n space cross times space 3.5 space R end fraction end style  ......................... (3)
 
For ideal gas , we have 
 
begin mathsize 14px style V space equals space open parentheses fraction numerator n space R over denominator P end fraction close parentheses space T end style
when pressure is kept constant , rate of change of volume is 
 
begin mathsize 14px style fraction numerator d V over denominator d T end fraction equals space open parentheses fraction numerator n R over denominator P end fraction close parentheses fraction numerator d T over denominator d t end fraction end style
 
Above expression can be written as
 
begin mathsize 14px style A space fraction numerator d h over denominator d t end fraction space equals space open parentheses fraction numerator n space R over denominator P end fraction close parentheses space fraction numerator d T over denominator d t end fraction end style ............................. (4) 
where A is area of cross section of cylinder and dh/dt is rate of vertical displacement of piston with respect to time , i.e, (dh/dt) is speed of piston .
 
From eqn.(3) and (4) , we get speed of piston as
 
begin mathsize 14px style fraction numerator d h over denominator d t end fraction space equals space open parentheses fraction numerator n cross times space R over denominator P cross times space A end fraction close parentheses space fraction numerator d T over denominator d t end fraction space equals space space open parentheses fraction numerator n cross times space R over denominator P cross times space A end fraction close parentheses space cross times fraction numerator 14 over denominator n space cross times space 3.5 space cross times R end fraction end style
begin mathsize 14px style fraction numerator d h over denominator d t end fraction space equals space fraction numerator 4 over denominator P space cross times A end fraction space equals space fraction numerator 4 over denominator 10 to the power of 5 cross times 10 cross times 10 to the power of negative 4 end exponent end fraction space equals space 0.04 space m divided by s space equals space 4 space c m divided by s end style
Speed of piston = 4 cm/s
 
Answered by Thiyagarajan K | 07 May, 2023, 09:54: PM
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