ThermodynamicsHard
Question
The internal energy of a monoatomic gas is 3 nRT . One mole of helium is kept in a cylinder having internal cross section area of $17{\text{ }cm}^{2}$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling 126 J heat. If the temperature rises by $4^{\circ}C$, then the piston will move $\_\_\_\_$ cm.
(atmospheric pressure $= 10^{5}\text{ }Pa$ )
Options
A.14.5
B.1.55
C.15.5
D.1.45
Solution
$\Delta U = 3nR\Delta T$
$\Delta U = 3 \times 1 \times \frac{25}{3} \times 4 = 100$ Joule
$${\Delta Q = 126 }{W = 26 = P\Delta V }{26 = 10^{5} \times 17 \times 10^{- 4}\Delta x }{\Delta x = \frac{26}{170} = 15.3}$$
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