At a pressure of `24 xx 10^(5) dyne cm^(-2)`. The volume of `O_(2)` is `10 litre` and mass is `20 g`. The rms velocity will be

At a pressure of 24xx10^(5) dyne/ cm^(2) ,the volume of O_(2) is 10 litre and mass is 20 gm.The r.m.s velocity will be

At a pressure of 24xx10^(5) dyne/ cm^(2) , the volume of N_(2) is 5litre and mass is 20gm. The 'rms' velocity will be (in m/sec)

A pressure of 10^(6)" dyne"//cm^(2) is equivalent to :

The volume of a gas at pressure 1.2 xx 10^(7) Nm^(-2) and temperature 127^(@)C is 2.0 litre. Find the number of molecules in the gas.

One gram molecule of nitrogen occupies 2 xx 10^(4) cm^(3) at a pressure of 10^(6) dyne cm ^(=-2). Given : N_(A)= 6 xx 10^(23). Which of the following is correct ?

The molecules of a given mass of gas have a rms velocity of 200 m//sec at 127^(@)C and 1.0 xx 10^(5) N//m^(2) pressure. When the temperature is 127^(@)C and pressure is 0.5 xx 10^(5) N//m^(2) the rms velocity in m//sec will be

Express a pressure of 2xx10^(5) pascal in units of (i) bar (ii) dyne cm^(-2) and (iii) millibar

A tank of capacity 20 litre contains air at a pressure of 4.0 xx 10^(5) Pa . Calculate the pressure that must be applied in order that the volume is reduced to 10 litres, keeping the temperature constant.