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 × 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:

Question

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 × 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: (A) 4 × 108S (B) 3 × 106S (C) 2 × 107S (D) 0.5 × 108S

Tardigrade Admin · Accepted Answer

t propto (Volume/velocity) vloume propto (T/P) ∴ t propto (√T/P) (t1/6 × 10-8) = (√5002P/) × (P/√300) t1 = 3.8 × 10-8 &asymp; 4 × 10-8

$4 \times 1 0^{8} S$

$3 \times 1 0^{6} S$

$2 \times 1 0^{7} S$

$0.5 \times 1 0^{8} S$

$t \propto \frac{V o l u m e}{v e l oc i t y}$
vloume $\propto \frac{T}{P}$
$∴ t$ \propto $\frac{T}{P}$
$\frac{t _{1}}{6 \times 1 0 ^{- 8}} = \frac{500}{2} P \times \frac{P}{300}$
$t_{1} = 3.8 \times 1 0^{- 8}$
≈ $4 \times 1 0^{- 8}$

Q. An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300K. The mean time between two successive collisions is 6×10−8s. If the pressure is doubled and temperature is increased to 500K, the mean time between two successive collisions will be close to:

4×108S

3×106S

2×107S

0.5×108S

t∝velocityVolume​ vloume ∝PT​ ∴t \propto PT​​ 6×10−8t1​​=2500​​P×300​P​ t1​=3.8×10−8 ≈ 4×10−8

$4 \times 1 0^{8} S$

$3 \times 1 0^{6} S$

$2 \times 1 0^{7} S$

$0.5 \times 1 0^{8} S$

$t \propto \frac{V o l u m e}{v e l oc i t y}$
vloume $\propto \frac{T}{P}$
$∴ t$ \propto $\frac{T}{P}$
$\frac{t _{1}}{6 \times 1 0 ^{- 8}} = \frac{500}{2} P \times \frac{P}{300}$
$t_{1} = 3.8 \times 1 0^{- 8}$
≈ $4 \times 1 0^{- 8}$