Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamicsMedium2 minQB

PhysicsMediumsingle choice

In an $H_{2}$ gas process, $P V^{2} =$ constant. The ratio of work done by gas to change in its internal energy is

Answer:

Solution:

$W = \frac{P _{1} V _{1} - P _{2} V _{2}}{2 - 1} = - n R Δ T; Δ U = \frac{n R Δ T}{γ - 1}$ with $γ = \frac{7}{5}$ .

$\frac{W}{Δ U} = - (γ - 1) = 1 - \frac{7}{5} = - \frac{2}{5} = - 0.4.$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics

⏱ 2mℹ️ Source: QB

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