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 $6 \times {10^{ - 8}}$ s. If the pressure is doubled and temperature is increased to 500 K, what will be the mean time between two successive collisions close to?
A. $4 \times {10^{ - 8}}$ s
B. $3 \times {10^{ - 6}}$ s
C. $2 \times {10^{ - 7}}$ s
D. $0.5 \times {10^{ - 8}}$ s

Answer 1

Hint: The mean time between two successive collisions is related to the volume and rms velocity of the gas. The rms velocity of the gas is given by the formula $\sqrt {\dfrac{{3RT}}{M}} $

Complete step by step answer:
The mean time between two successive collisions varies inversely with the velocity of the gas molecules. It is directly proportional to the volume of the gas. But, we do not know of any direct relationship between pressure, temperature and mean time between two successive collisions. The relation between time and volume and velocity can be written as,
t $\alpha \dfrac{V}{v}$ …equation (1)
where V is the volume and v is the rms velocity of the gas molecules.
Since the gas is ideal, the ideal gas equation can be applied on the system. Using the ideal gas equation, we can find out the expression of volume in terms of temperature and pressure. We obtain,
$PV = nRT \Rightarrow V = \dfrac{{nRT}}{P}$ …equation (2)
We have no information in the question about the velocity of the gas molecules in initial and final state. To write the expression of velocity in terms of temperature, we use the formula for the rms velocity of the gas molecules, which is,
$v = \sqrt {\dfrac{{3RT}}{M}} $ …equation (3)
On substituting the expressions obtained in equations (2) and (3) into equation (1), we obtain,
t $\alpha \dfrac{{\dfrac{{nRT}}{P}}}{{\sqrt {\dfrac{{3RT}}{M}} }}\alpha \dfrac{{\sqrt T }}{P}$
Let us denote the values of pressure and temperature at the initial state with the subscript 1 and that at the final state by the subscript 2. The time between two collisions is given and we need to find out the time at the final state. We can find out the time between two collisions at the final state by taking the ratio as follows,
$\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{\dfrac{{\sqrt {{T_2}} }}{{{P_2}}}}}{{\dfrac{{\sqrt {{T_1}} }}{{{P_1}}}}} = \dfrac{{\sqrt {{T_2}} }}{{{P_2}}} \times \dfrac{{{P_1}}}{{\sqrt {{T_1}} }} = \dfrac{{\sqrt {500} }}{{2{P_1}}} \times \dfrac{{{P_1}}}{{\sqrt {300} }} = \dfrac{{\sqrt 5 }}{{2\sqrt 3 }} = 0.645$
$ \Rightarrow {t_2} = 0.645 \times {t_1} = 0.645 \times 6 \times {10^{ - 8}} \approx 4 \times {10^{ - 8}}$ s
Hence, the mean time between two successive collisions at the final state is $4 \times {10^{ - 8}}$ s.
Therefore, the correct answer is option A.

Note:The gas of the system remains the same in both the cases. Hence, in expression for the mean time between two collisions, molecular mass of the gas can be treated as a constant in this question. However, if the gas is also changed in the second case, the mean time between two collisions will also depend on the square root of the mass of the gas in the second case.