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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermodynamicsMedium2 minQB

PhysicsMediumsingle choice

A process $1 \to 2$ using monoatomic gas is shown on the P-V diagram. $P_{1} = 2 P_{2} = 1 0^{6}$ $N / m^{2}, V_{2} = 4 V_{1} = 0.4 m^{3}$ . The heat absorbed by the gas in this process will be:

Options:

Answer:

Solution:

$W =$ area under the straight line $= \frac{P _{1} + P _{2}}{2} (V_{2} - V_{1}) = \frac{1 0 ^{6} + 0.5 \times 1 0 ^{6}}{2} \times 0.3 = 2.25 \times 1 0^{5} J$ .

$Δ U = \frac{3}{2} (P_{2} V_{2} - P_{1} V_{1}) = \frac{3}{2} (0.5 \times 1 0^{6} \times 0.4 - 1 0^{6} \times 0.1) = 1.5 \times 1 0^{5} J . Q = Δ U + W = 3.75 \times 1 0^{5} J = 375 kJ .$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics

⏱ 2mℹ️ Source: QB

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