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Heat and Thermodynamics question

2008 · Q158
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Heat and Thermodynamics question

2008 · Q158

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
At 10oC the value of the density of a fixed mass of an ideal gas divided by it pressure is x. At 110oC this ratio is
  1. A
    10110x{{10} \over {110}}x11010​x
  2. B
    283383x{{283} \over {383}}x383283​x
  3. C
    xxx
  4. D
    383283x{{383} \over {283}}x283383​x
View written solutionFree

Correct answer: B

PV = nRT

⇒PV=mMRT \Rightarrow PV = {m \over M}RT⇒PV=Mm​RT

⇒PVm=RTM \Rightarrow {{PV} \over m} = {{RT} \over M}⇒mPV​=MRT​

⇒Pρ=RTM \Rightarrow {P \over \rho } = {{RT} \over M}⇒ρP​=MRT​

⇒Pρ∝T \Rightarrow {P \over \rho } \propto T⇒ρP​∝T

∴Pρ∝1T \therefore {P \over \rho } \propto {1 \over T}∴ρP​∝T1​

P1ρ1P2ρ2=T1T2=383283{{{{{P_1}} \over {{\rho _1}}}} \over {{{{P_2}} \over {{\rho _2}}}}} = {{{T_1}} \over {{T_2}}} = {{383} \over {283}}ρ2​P2​​ρ1​P1​​​=T2​T1​​=283383​

⇒xP2P2=383283 \Rightarrow {x \over {{{{P_2}} \over {{P_2}}}}} = {{383} \over {283}}⇒P2​P2​​x​=283383​

∴P2P2=383283x \therefore {{{P_2}} \over {{P_2}}} = {{383} \over {283}}x∴P2​P2​​=283383​x

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