A sample of gas occupies 21 L under a pressure of 1.3 atm. What would the resulting volume be if the pressure were increased to 3.9 atm if the temperature did not change?

First look at the things that need to be converted $35.9 ^@"C" = "(35.9 + 273.15) K" = "309.0 K"$ $"25.5 kPa = 0.2467 atm"$ $"500 mL = 0.500 L"$...

Given $Pprop1/V$, $P_1V_1=P_2V_2$, and the great advantage of this proportionality is that we can use unorthodox units of pressure and volume, i.e. $"psi, pints, atmosphere, gallons"$ so long as we...

This is an example of a problem. $color(blue)(|bar(ul((P_1V_1)/T_1 = (P_2V_2)/T_2|)$ We can rearrange this formula to get $V_2 = V_1 × P_1/P_2 × T_2/T_1$...

And so.................. $V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)$ $=(2.31*atmxx17.5*Lxx350*K)/(299*Kxx1.75*atm)<=30*L.$

And $1*"bar"-=(1*"bar")/(1.01325*atm*"bar"^-1)=0.987*atm$ I do all this rigmarole BECAUSE I know that $1*atm$ will support a column of mercury that is $760*mm$ high........... And hence in terms of the length of...

holds that $P_1V_1=P_2V_2$ at constant temperature. And thus $P_2=(P_1V_1)/V_2=(1.2*atmxx8*mL)/(12*mL)=0.8*atm$. This decrease in pressure is reasonable, given that we have INCREASED the volume.

The idea here is that the volume and the temperature of a gas have a direct relationship when the pressure and the number of moles of gas are being kept...

$(P_1V_1)/T_1=(P_2V_2)/T_2$ You need to convert degC to Kelvin by adding 273: $:.(200xx25)/298=(250xxV_2)/273$ $:.V_2=(200xx25xx273)/(298xx250)$ $:.V_2=18.32"L"$

When you are given this much information in the context of a generic, unnamed gas, it's a good idea to consider the : $PV = nRT$ Before we...

Even without doing any calculations, you should be able to look at the information given and predict that the volume of the gas will increase after temperature is increased and...