An ideal gas is expanding such that PT 2 = constant. The coefficient of volume expansion of gas is T A where T is temperature in Kelvin. Find the value of A.

PT 2 = K ⇒ ( V nRT ) T 2 = K ⇒ V = K ′ T 3 ..... (i) dt dv = K ′ 3 T 2 ..... (ii) ; γ = V 1 ( dt dv ) from (i) & (ii): γ = T 3 k ′ × K ′ 3 T 2 = T 3 ⇒ A = 3

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PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

PhysicsMediuminteger

An ideal gas is expanding such that $PT^{2} =$ constant. The coefficient of volume expansion of gas is $\frac{A}{T}$ where T is temperature in Kelvin. Find the value of A.

Answer:

Solution:

$PT^{2} = K \Rightarrow (\frac{nRT}{V}) T^{2} = K \Rightarrow V = \frac{T ^{3}}{K ^{'}}$ .....

(i)

$\frac{dv}{dt} = \frac{3 T ^{2}}{K ^{'}}$ .....

(ii) ; $γ = \frac{1}{V} (\frac{dv}{dt})$

from (i) & (ii): $γ = \frac{k ^{'}}{T ^{3}} \times \frac{3 T ^{2}}{K ^{'}} = \frac{3}{T} \Rightarrow A = 3$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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