A gas that has a volume of 28L a temp of 45 degrees and a unknown pressure has its volume increases to 34 L and its temp. decreased to 35 degrees the pressure after the change is measured and found to be 2.0atm. what was the original pressure of the gas?

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Answer 1 · 2018-05-16T14:49:25

Well, the combined gas equation holds that....

#(P_1V_1)/T_1=(P_2V_2)/T_2#...where we have #"la temperatura assoluta"#...

#=""^@C+273.15*K#

And so #P_1=(P_2V_2)/T_2xxT_1/V_1#

#=(2.0*atmxx34*L)/(308.15*K)xx(318.15*K)/(28*L)=??*atm#