If the oxygen is replaced by helium at the same temperature, how many kilograms of the latter will be needed to produce a gauge pressure of 5.40 atm?

This is an incomplete question, here is a complete question.

A tank contains 30.0 kg of O₂ gas at gauge pressure of 8.20 atm. If the oxygen is replaced by helium at the same temperature. How many kilograms of the latter will be needed to produce a gauge pressure of 7.00 atm?

Answer : The mass of latter needed will be, 2.47 kg

Explanation :

Using ideal gas equation:

where,

P = pressure of gas

V = volume of gas

T = temperature of gas

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas

M = molar mass of gas

The expression for two gases at same temperature, volume of gas will be:

where,

= pressure of O₂ gas = 8.20 atm

= pressure of He gas = 5.40 atm

= molar mass of O₂ gas = 32 g/mol

= molar mass of He gas = 4 g/mol

= mass of O₂ gas = 30.0 kg

= mass of He gas = ?

Now put all the given values in the ideal gas equation, we get:

Therefore, the mass of latter needed will be, 2.47 kg

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation:

this is poiselle's equation from fluid mechanic in physics. mech eng