An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as P = Pₒ(1 + (Vₒ/V)²)⁻¹, where Pₒ and Vₒ are constants. If two samples A and B (two moles each) with initial volumes Vₒ and 3Vₒ respectively undergo the above process and attain the same pressure, then the di…

Question

An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as P = Pₒ(1 + (Vₒ/V)²)⁻¹, where Pₒ and Vₒ are constants. If two samples A and B (two moles each) with initial volumes Vₒ and 3Vₒ respectively undergo the above process and attain the same pressure, then the difference at the temperatures of these samples, T_B-T_A is . (R = gas constant)

Accepted Answer

(2) (11PₒVₒ)/(10R). The law is P=(P₀)/(1+((V₀)/V)²), so the pressure is fixed once the volume is. Sample A at V=V₀: P_A=(P₀)/(1+1)=(P₀)/2. Sample B at V=3V₀: P_B=(P₀)/(1+1/9)=(9P₀)/(10). With n=2 for each, T=(PV)/(nR). T_A=((P₀/2)V₀)/(2R)=(P₀V₀)/(4R) and T_B=((9P₀/10)(3V₀))/(2R)=(27P₀V₀)/(20R). T_B-T_A=(27P₀V₀)/(20R)-(5P₀V₀)/(20R)=(11P₀V₀)/(10R). (The stem's phrase 'attain the same pressure' cannot hold for both samples on this law; the difference above is built from the two initial states. See DEFECTS.md.)

Step-by-step solution

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