At 25°C and 730 mm pressure,380 ml of dry oxygen was collected.If the temperature is constant ,what volume will the oxygen occupy at 760 mm pressure ?

ANSWER- 365ml

p1v1=p2v2

730*380/760=v2
365ml

Do you want to know if your answer of 365 is correct? It is. Do you want to know how to work it?

P1V1 = P2V2

P1v1= p2v2 730×380=760× v2 277400=760v2 v2 =277400/760=365 ml

P1v1=p2v2

730*380 = 760*v2
V2=365ml

730*380/760=365

At hundred degree Celsius and 1 ATM pressure if the density of liquid water is 1 gram per cm cube and that of water vapour is 0.0006 gram per centimetre cube the volume occupied by the water molecule in 1 litre of steam at the temperature is

To solve this problem, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure when the temperature is constant.

We have the initial volume (380 ml) and pressure (730 mm) and need to find the final volume at a different pressure (760 mm).

Boyle's Law can be stated as:

P1 * V1 = P2 * V2

Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

In this case, we know P1 = 730 mm, V1 = 380 ml, and P2 = 760 mm. We need to solve for V2.

Using the formula, we have:

730 mm * 380 ml = 760 mm * V2

Now we can solve for V2:

V2 = (730 mm * 380 ml) / 760 mm

V2 = 365 ml

Therefore, the volume of oxygen at 760 mm pressure would be 365 ml.