ThermodynamicsMedium
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
$${\Delta Q = 126 }{W = 26 = P\Delta V }{26 = 10^{5} \times 17 \times 10^{- 4}\Delta x }{\Delta x = \frac{26}{170} = 15.3}$$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
Which one of the following is not thermodynamical co-ordinate ?...Two moles of ideal helium gas are in a rubber balloon at 30oC. The balloon is fully expandable and can be assumed to req...For an ideal gas, the initial pressure and volume are equal to the final pressure and volume....Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1, T2), volume (V1, V2) and pressure (P1, ...A gas kept in a container, if the container is of finite conductivity, then the process...