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Heat and Thermodynamics question

2023 · 30 Jan · Shift 1 · Q44
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  5. /2023 · 30 Jan · Shift 1 · Q44

Heat and Thermodynamics question

2023 · 30 Jan · Shift 1 · Q44

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The pressure (P)(\mathrm{P})(P) and temperature (T)\mathrm{T})T) relationship of an ideal gas obeys the equation PT2=\mathrm{PT}^{2}=PT2= constant. The volume expansion coefficient of the gas will be :
  1. A
    3T23 T^{2}3T2
  2. B
    3T2\frac{3}{T^2}T23​
  3. C
    3T3\frac{3}{T^3}T33​
  4. D
    3T\frac{3}{T}T3​
View written solutionFree

Correct answer: D

  1. Use the ideal gas equation

For an ideal gas, PV=nRTPV = nRTPV=nRT where nRnRnR is constant for a fixed amount of gas.

  1. Given relation between pressure and temperature

The problem states: PT2=constantPT^2 = \text{constant}PT2=constant So, P∝1T2P \propto \frac{1}{T^2}P∝T21​

Let P=kT2P = \frac{k}{T^2}P=T2k​ for some constant kkk.

  1. Find how volume depends on temperature

From the ideal gas law, V=nRTPV = \frac{nRT}{P}V=PnRT​ Substitute P=kT2P = \frac{k}{T^2}P=T2k​: V=nRTk/T2=nRkT3V = \frac{nRT}{k/T^2} = \frac{nR}{k}T^3V=k/T2nRT​=knR​T3 Hence, V∝T3V \propto T^3V∝T3

  1. Volume expansion coefficient

The coefficient of volume expansion is defined as β=1VdVdT\beta = \frac{1}{V}\frac{dV}{dT}β=V1​dTdV​

Since V=cT3V = cT^3V=cT3 for some constant ccc, we get dVdT=3cT2\frac{dV}{dT} = 3cT^2dTdV​=3cT2 Therefore, β=1cT3(3cT2)=3T\beta = \frac{1}{cT^3}(3cT^2) = \frac{3}{T}β=cT31​(3cT2)=T3​

  1. Match with the options

Thus, the volume expansion coefficient is 3T\boxed{\frac{3}{T}}T3​​ So the correct option is D.

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