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Heat and Thermodynamics question

2019 · 10 Apr · Shift 2 · Q58
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Heat and Thermodynamics question

2019 · 10 Apr · Shift 2 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
One mole of ideal gas passes through a process where pressure and volume obey the relation P=P0[1−12(V0V)2]P = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]P=P0​[1−21​(VV0​​)2]. Here P0 and V0 are constants. Calculate the change in the temperature of the gas if its volume changes form V0 to 2V0
  1. A
    34P0V0R{3 \over 4}{{{P_0}{V_0}} \over R}43​RP0​V0​​
  2. B
    12P0V0R{1 \over 2}{{{P_0}{V_0}} \over R}21​RP0​V0​​
  3. C
    54P0V0R{5 \over 4}{{{P_0}{V_0}} \over R}45​RP0​V0​​
  4. D
    14P0V0R{1 \over 4}{{{P_0}{V_0}} \over R}41​RP0​V0​​
View written solutionFree

Correct answer: C

  1. Use the ideal gas equation

For one mole of an ideal gas, PV=RTPV = RTPV=RT So, T=PVRT = \frac{PV}{R}T=RPV​

Given the process relation, P=P0[1−12(V0V)2]P = P_0\left[1 - \frac{1}{2}\left(\frac{V_0}{V}\right)^2\right]P=P0​[1−21​(VV0​​)2]

Hence, T(V)=VR⋅P0[1−12(V0V)2]T(V) = \frac{V}{R} \cdot P_0\left[1 - \frac{1}{2}\left(\frac{V_0}{V}\right)^2\right]T(V)=RV​⋅P0​[1−21​(VV0​​)2]

  1. Simplify the expression for temperature

T(V)=P0R[V−12V(V02V2)]T(V) = \frac{P_0}{R}\left[V - \frac{1}{2}V\left(\frac{V_0^2}{V^2}\right)\right]T(V)=RP0​​[V−21​V(V2V02​​)]

T(V)=P0R[V−12V02V]T(V) = \frac{P_0}{R}\left[V - \frac{1}{2}\frac{V_0^2}{V}\right]T(V)=RP0​​[V−21​VV02​​]

  1. Initial temperature at V=V0V=V_0V=V0​

T1=P0R[V0−12V02V0]T_1 = \frac{P_0}{R}\left[V_0 - \frac{1}{2}\frac{V_0^2}{V_0}\right]T1​=RP0​​[V0​−21​V0​V02​​]

T1=P0R[V0−12V0]T_1 = \frac{P_0}{R}\left[V_0 - \frac{1}{2}V_0\right]T1​=RP0​​[V0​−21​V0​]

T1=P0V02RT_1 = \frac{P_0V_0}{2R}T1​=2RP0​V0​​

  1. Final temperature at V=2V0V=2V_0V=2V0​

T2=P0R[2V0−12V022V0]T_2 = \frac{P_0}{R}\left[2V_0 - \frac{1}{2}\frac{V_0^2}{2V_0}\right]T2​=RP0​​[2V0​−21​2V0​V02​​]

T2=P0R[2V0−V04]T_2 = \frac{P_0}{R}\left[2V_0 - \frac{V_0}{4}\right]T2​=RP0​​[2V0​−4V0​​]

T2=P0R⋅7V04=7P0V04RT_2 = \frac{P_0}{R}\cdot \frac{7V_0}{4} = \frac{7P_0V_0}{4R}T2​=RP0​​⋅47V0​​=4R7P0​V0​​

  1. Change in temperature

ΔT=T2−T1\Delta T = T_2 - T_1ΔT=T2​−T1​

ΔT=7P0V04R−P0V02R\Delta T = \frac{7P_0V_0}{4R} - \frac{P_0V_0}{2R}ΔT=4R7P0​V0​​−2RP0​V0​​

ΔT=7P0V04R−2P0V04R\Delta T = \frac{7P_0V_0}{4R} - \frac{2P_0V_0}{4R}ΔT=4R7P0​V0​​−4R2P0​V0​​

ΔT=5P0V04R\Delta T = \frac{5P_0V_0}{4R}ΔT=4R5P0​V0​​

  1. Match with options

ΔT=54P0V0R\Delta T = \frac{5}{4}\frac{P_0V_0}{R}ΔT=45​RP0​V0​​

So the correct option is C.

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