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Heat and Thermodynamics question

2021 · 27 Aug · Shift 1 · Q50
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Heat and Thermodynamics question

2021 · 27 Aug · Shift 1 · Q50

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
An ideal gas is expanding such that PT3 = constant. The coefficient of volume expansion of the gas is :
  1. A
    1T{1 \over T}T1​
  2. B
    2T{2 \over T}T2​
  3. C
    4T{4 \over T}T4​
  4. D
    3T{3 \over T}T3​
View written solutionFree

Correct answer: C

  1. Given process condition

The gas expands such that PT3=constant.PT^3=\text{constant}.PT3=constant.

For an ideal gas, PV=nRT.PV=nRT.PV=nRT.

We need the coefficient of volume expansion: β=1V(∂V∂T).\beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right).β=V1​(∂T∂V​).

Since the process relation connects PPP and TTT, we first express VVV in terms of TTT.

  1. Express pressure in terms of temperature

From PT3=kPT^3 = kPT3=k (where kkk is a constant), P=kT3.P = \frac{k}{T^3}.P=T3k​.

  1. Use ideal gas equation to find V(T)V(T)V(T)

Using PV=nRT,PV=nRT,PV=nRT, we get V=nRTP.V = \frac{nRT}{P}.V=PnRT​.

Substitute P=kT3P=\dfrac{k}{T^3}P=T3k​: V=nRTk/T3=nRkT4.V = \frac{nRT}{k/T^3} = \frac{nR}{k}T^4.V=k/T3nRT​=knR​T4.

So, V∝T4.V \propto T^4.V∝T4.

  1. Differentiate to find coefficient of volume expansion

Let V=cT4V = cT^4V=cT4 for some constant ccc.

Then, dVdT=4cT3.\frac{dV}{dT} = 4cT^3.dTdV​=4cT3.

Therefore, β=1VdVdT=1cT4(4cT3)=4T.\beta = \frac{1}{V}\frac{dV}{dT} = \frac{1}{cT^4}(4cT^3)=\frac{4}{T}.β=V1​dTdV​=cT41​(4cT3)=T4​.

  1. Match with the options

β=4T\beta = \frac{4}{T}β=T4​ which corresponds to Option C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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