JEE Mains · Physics · STD 11 - 11. thermodynamics
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 \(4\) cm\(^2\). The gas is heated slowly to raise the temperature by \(1.2\,^\circ\)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)
- A \(24.8\)
- B \(25\)
- C \(15.04\)
- D \(29.98\)
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
Using the first law of thermodynamics: \(Q = \Delta U + W\) Since the gas is diatomic with rotational modes only, the degrees of freedom are \(f = 5\), so: \(C_V = \dfrac{5}{2}R\) For \(n = 1\text{ mole}\) and \(\Delta T = 1.2\text{ K}\):…
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