One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is 4 cm2. The gas is heated slowly to raise the temperature by 1.2 °C during which the piston moves by 25 mm. The amount of heat supplied to the gas is J. (Atmospheric pressure = 100 kPa, R=8.3 J/mol. K) (Neglect mass of the piston)
- 24.8
- 10.96
- 15.04
- 29.98
Answer
(2) 10.96
Solution
Rotational modes only: a diatomic molecule has 2 rotational degrees of freedom f=2
CV= f2R=R
U=nCV T=1 8.3 1.2=9.96 J
Massless piston moving slowly gas pressure = atmospheric pressure (isobaric)
W=P A x=105 (4 10-4) (25 10-3)=1.0 J
First law: Q= U+W=9.96+1.0=10.96 J
Answer: (2)