Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoFirst lawMedium2 minPYQ_2024

PhysicsMediumsingle choice

A total of $48 J$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} C$ . The work done by the gas is: Given, $R = 8.3 J K^{- 1} mol^{- 1}$ .

Options:

Answer:

Solution:

$1^{st}$ law of thermodynamics $Δ Q = Δ U + W \Rightarrow + 48 = nC_{v} Δ T + W \Rightarrow 48 = (1) (\frac{3 R}{2}) (2) + W \Rightarrow W = 48 - 3 \times R \Rightarrow W = 48 - 3 \times (8.3) \Rightarrow W = 23.1 Joule$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:First law

⏱ 2mℹ️ Source: PYQ_2024

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