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Heat and Thermodynamics question

2011 · Q100
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Heat and Thermodynamics question

2011 · Q100

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
A mass of diatomic gas (γ=1.4)(\gamma = 1.4)(γ=1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27oC to 927oC. The pressure of the gas in the final state is
  1. A
    8 atm
  2. B
    28 atm
  3. C
    68.7 atm
  4. D
    256 atm
View written solutionFree

Correct answer: D

T1 = 273 + 27 = 300K

T2 = 273 + 927 = 1200K

For adiabatic process,

P1−γTγ{P^{1 - \gamma }}{T^\gamma }P1−γTγ = constant

⇒P11−γT1γ=P21−γT2γ\Rightarrow {P_1}^{1 - \gamma }{T_1}^\gamma = {P_2}^{1 - \gamma }{T_2}^\gamma⇒P1​1−γT1​γ=P2​1−γT2​γ

⇒(P2P1)1−γ=(T1T2)γ \Rightarrow {\left( {{{{P_2}} \over {{P_1}}}} \right)^{1 - \gamma }} = {\left( {{{{T_1}} \over {{T_2}}}} \right)^\gamma }⇒(P1​P2​​)1−γ=(T2​T1​​)γ

⇒(P1T2)1−γ=(T2T1)γ \Rightarrow {\left( {{{{P_1}} \over {{T_2}}}} \right)^{1 - \gamma }} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^\gamma }⇒(T2​P1​​)1−γ=(T1​T2​​)γ

(P1P2)1−1.4=(1200300)1.4{\left( {{{{P_1}} \over {{P_2}}}} \right)^{1 - 1.4}} = {\left( {{{1200} \over {300}}} \right)^{1.4}}(P2​P1​​)1−1.4=(3001200​)1.4

(P1P2)−0.4=(4)1.4{\left( {{{{P_1}} \over {{P_2}}}} \right)^{ - 0.4}} = {\left( 4 \right)^{1.4}}(P2​P1​​)−0.4=(4)1.4

(P2P1)0.4=41.4{\left( {{{{P_2}} \over {{P_1}}}} \right)^{0.4}} = {4^{1.4}}(P1​P2​​)0.4=41.4

P2=P14(1.40.4)=P14(72){P_2} = {P_1}{4^{\left( {{{1.4} \over {0.4}}} \right)}} = {P_1}{4^{\left( {{7 \over 2}} \right)}}P2​=P1​4(0.41.4​)=P1​4(27​)

= P1 (27) = 2 × 128 = 256 atm

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