A sample of gas has a volume of 20 cm³.The pressure is changed to 90 kPa at constant temperature, while the volume increases to 75 cm³.What was the original pressure of the gas?​

337.5kPa ~ 338kPa

Explanation:

Using the ideal gas law PV=nRT we have the following definitions from the problem:

V(initial) = 20cm³

P(initial) = ?kPa

V(final) = 75cm³

P(final) = 90kPa

Since we know that the number of moles of the sample did not change, nor did the temperature, nor does the ideal gas constant (R) we can rewrite this equation to state:

P(initial)V(initial) = nRT =P(final)V(final) ~  P(initial)V(initial) = P(final)V(final)

Rearranging this equation as we are solving for the initial pressure we find that:

P(initial) = (P(final)V(final))/V(initial)

P(initial) = ((90kPa)(75cm³))/20cm³

P(initial) = 337.5kPa ~ 338kPA

not clear

the answer for sodium is 0.186

the answer for rubidium is 0.251