Thermodynamics

JEE Main level, single correct, medium

61 of 190

Question 61 of 190

An ideal gas undergoes a process maintaining relation between pressure (P)(P) and volume (V)(V) as P=Po(1+(VoV)2)−1P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}, where PoP_o and VoV_o are constants. If two samples AA and BB (two moles each) with initial volumes VoV_o and 3Vo3V_o respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, TB−TAT_B - T_A is _____. (R=R = gas constant)

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Answer: (B) 11PoVo10R\dfrac{11P_o V_o}{10R}

The given relation between pressure and volume is:

P=Po(1+(VoV)2)−1P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}

For sample A, the initial volume is VA=VoV_A = V_o. Substituting this into the given relation, the initial pressure PAP_A is:

PA=Po(1+(VoVo)2)−1=Po2P_A = P_o\left(1 + \left(\dfrac{V_o}{V_o}\right)^2\right)^{-1} = \dfrac{P_o}{2}

Using the ideal gas equation PV=nRTPV = nRT for n=2n = 2 moles, the initial temperature TAT_A is:

TA=PAVAnR=(Po2)Vo2R=PoVo4RT_A = \dfrac{P_A V_A}{nR} = \dfrac{\left(\dfrac{P_o}{2}\right) V_o}{2R} = \dfrac{P_o V_o}{4R}

For sample B, the initial volume is VB=3VoV_B = 3V_o. Substituting this into the relation, the initial pressure PBP_B is:

PB=Po(1+(Vo3Vo)2)−1=Po(1+19)−1=9Po10P_B = P_o\left(1 + \left(\dfrac{V_o}{3V_o}\right)^2\right)^{-1} = P_o\left(1 + \dfrac{1}{9}\right)^{-1} = \dfrac{9P_o}{10}

The initial temperature TBT_B for n=2n = 2 moles is:

TB=PBVBnR=(9Po10)(3Vo)2R=27PoVo20RT_B = \dfrac{P_B V_B}{nR} = \dfrac{\left(\dfrac{9P_o}{10}\right) (3V_o)}{2R} = \dfrac{27P_o V_o}{20R}

The difference in the initial temperatures of the two samples is:

TB−TA=27PoVo20R−PoVo4RT_B - T_A = \dfrac{27P_o V_o}{20R} - \dfrac{P_o V_o}{4R}

TB−TA=27PoVo−5PoVo20R=22PoVo20R=11PoVo10RT_B - T_A = \dfrac{27P_o V_o - 5P_o V_o}{20R} = \dfrac{22P_o V_o}{20R} = \dfrac{11P_o V_o}{10R}

Answer: 11PoVo10R\dfrac{11P_o V_o}{10R}

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