3 mol of a gas with C p = 7 cal m o l − 1 K − 1 is heated at constant volume from 1 0 ∘ C to 3 0 ∘ C. Take R = 2 cal m o l − 1 K − 1 . Find the heat absorbed in calories.

C v = 7 − 2 = 5 . Q = 3 × 5 × 20 = 300 cal.

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PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesEasy2 minQB

PhysicsEasyinteger

3 mol of a gas with $C_{p} = 7$ cal $m o l^{- 1}$ $K^{- 1}$ is heated at constant volume from $1 0^{\circ}$ C to $3 0^{\circ}$ C. Take $R = 2$ cal $m o l^{- 1}$ $K^{- 1}$ . Find the heat absorbed in calories.

Answer:

300

Solution:

$C_{v} = 7 - 2 = 5$ . $Q = 3 \times 5 \times 20 = 300$ cal.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

⏱ 2mℹ️ Source: QB

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