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NEET PHYSICSEasy

At 10C10^\circ\text{C} the value of the density of a fixed mass of an ideal gas divided by its pressure is xx. At 110C110^\circ\text{C} this ratio is:

A

xx

B

383283x\frac{383}{283}x

C

10110x\frac{10}{110}x

D

283383x\frac{283}{383}x

Step-by-Step Solution

According to the Ideal Gas Equation, PV=nRTPV = nRT. Since n=mMn = \frac{m}{M} (mass / Molar mass) and density ρ=mV\rho = \frac{m}{V}, we can substitute to find the relationship between density and pressure: P=mVRTM=ρRTMP = \frac{m}{V} \frac{RT}{M} = \rho \frac{RT}{M} Rearranging for the ratio of density to pressure: ρP=MRT\frac{\rho}{P} = \frac{M}{RT} For a fixed mass of a specific gas, the Molar mass (MM) is constant. Thus, the ratio is inversely proportional to the absolute temperature: ρP1T\frac{\rho}{P} \propto \frac{1}{T}

  1. Initial Condition: T1=10C=10+273=283 KT_1 = 10^\circ\text{C} = 10 + 273 = 283 \text{ K}. x=k283(where k is a constant)x = \frac{k}{283} \quad \text{(where } k \text{ is a constant)}
  2. Final Condition: T2=110C=110+273=383 KT_2 = 110^\circ\text{C} = 110 + 273 = 383 \text{ K}. x=k383x' = \frac{k}{383}
  3. Calculate Ratio: xx=k/383k/283=283383\frac{x'}{x} = \frac{k / 383}{k / 283} = \frac{283}{383} x=283383xx' = \frac{283}{383}x
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