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Heat and Thermodynamics question

2019 · 12 Jan · Shift 2 · Q67
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Heat and Thermodynamics question

2019 · 12 Jan · Shift 2 · Q67

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 ×\times× 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to
  1. A
    0.5 ×\times× 10 −-− 8 s
  2. B
    4 ×\times× 10 −-− 8 s
  3. C
    3 ×\times× 10 −-− 6 s
  4. D
    2 ×\times× 10 −-− 7 s
View written solutionFree

Correct answer: B

  1. Relation for collision time

For an ideal gas, the mean time between successive collisions is

τ=λvˉ\tau = \frac{\lambda}{\bar v}τ=vˉλ​

where:

  • λ\lambdaλ = mean free path
  • vˉ\bar vvˉ = average molecular speed

Now,

λ∝1n\lambda \propto \frac{1}{n}λ∝n1​

for a given gas, where nnn is number density.

Using ideal gas law,

n∝PTn \propto \frac{P}{T}n∝TP​

so

λ∝TP\lambda \propto \frac{T}{P}λ∝PT​

Also, molecular speed varies as

vˉ∝T\bar v \propto \sqrt{T}vˉ∝T​

Therefore,

τ=λvˉ∝T/PT=TP\tau = \frac{\lambda}{\bar v} \propto \frac{T/P}{\sqrt{T}} = \frac{\sqrt{T}}{P}τ=vˉλ​∝T​T/P​=PT​​

So,

τ∝TP\tau \propto \frac{\sqrt{T}}{P}τ∝PT​​
  1. Apply the proportionality

Initial conditions:

P1=2 atm,T1=300 K,τ1=6×10−8 sP_1 = 2\,\text{atm}, \quad T_1 = 300\,\text{K}, \quad \tau_1 = 6 \times 10^{-8}\,\text{s}P1​=2atm,T1​=300K,τ1​=6×10−8s

Final conditions:

P2=4 atm,T2=500 KP_2 = 4\,\text{atm}, \quad T_2 = 500\,\text{K}P2​=4atm,T2​=500K

Hence,

τ2τ1=T2/P2T1/P1\frac{\tau_2}{\tau_1} = \frac{\sqrt{T_2}/P_2}{\sqrt{T_1}/P_1}τ1​τ2​​=T1​​/P1​T2​​/P2​​

Substitute values:

τ26×10−8=500/4300/2\frac{\tau_2}{6 \times 10^{-8}} = \frac{\sqrt{500}/4}{\sqrt{300}/2}6×10−8τ2​​=300​/2500​/4​ =5004⋅2300=12500300= \frac{\sqrt{500}}{4} \cdot \frac{2}{\sqrt{300}} = \frac{1}{2}\sqrt{\frac{500}{300}}=4500​​⋅300​2​=21​300500​​ =1253= \frac{1}{2}\sqrt{\frac{5}{3}}=21​35​​

Now,

53≈1.29\sqrt{\frac{5}{3}} \approx 1.2935​​≈1.29

so

τ2τ1≈1.292≈0.645\frac{\tau_2}{\tau_1} \approx \frac{1.29}{2} \approx 0.645τ1​τ2​​≈21.29​≈0.645

Thus,

τ2≈6×10−8×0.645\tau_2 \approx 6 \times 10^{-8} \times 0.645τ2​≈6×10−8×0.645 τ2≈3.87×10−8 s\tau_2 \approx 3.87 \times 10^{-8}\,\text{s}τ2​≈3.87×10−8s

This is closest to

4×10−8 s4 \times 10^{-8}\,\text{s}4×10−8s
  1. Evaluate options
  • A: 0.5×10−80.5 \times 10^{-8}0.5×10−8 s = too small
  • B: 4×10−84 \times 10^{-8}4×10−8 s = matches
  • C: 3×10−63 \times 10^{-6}3×10−6 s = much too large
  • D: 2×10−72 \times 10^{-7}2×10−7 s = too large

Therefore, the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.

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