JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 23)
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
0.5 $$ \times $$ 10$$-$$8 s
4 $$ \times $$ 10$$-$$8 s
3 $$ \times $$ 10$$-$$6 s
2 $$ \times $$ 10$$-$$7 s
Explanation
t $$ \propto $$ $${{Volume} \over {velocity}}$$
volume $$ \propto $$ $${T \over P}$$
$$ \therefore $$ t $$ \propto $$ $${{\sqrt T } \over P}$$
$${{{t_1}} \over {6 \times {{10}^{ - 8}}}} = {{\sqrt {500} } \over {2P}} \times {P \over {\sqrt {300} }}$$
t1 = 3.8 $$ \times $$ 10$$-$$8
$$ \approx $$ 4 $$ \times $$ 10$$-$$8
volume $$ \propto $$ $${T \over P}$$
$$ \therefore $$ t $$ \propto $$ $${{\sqrt T } \over P}$$
$${{{t_1}} \over {6 \times {{10}^{ - 8}}}} = {{\sqrt {500} } \over {2P}} \times {P \over {\sqrt {300} }}$$
t1 = 3.8 $$ \times $$ 10$$-$$8
$$ \approx $$ 4 $$ \times $$ 10$$-$$8
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