A gas that has a volume of 28 liters, a temperature of 42°C, and an unknown pressure has its volume increased to 49 liters and its temperature decreased to 27°C. If the pressure is measured after the change at 4.0 atm, what was the original pressure of the gas?

7.35atm

Explanation:

Data obtained from the question include:

V1 = 28L

T1 = 42°C = 42 + 273 = 315K

P1 =?

V2 = 49L

T2 = 27°C = 27 + 273 = 300K

P2 = 4atm

Using P1V1/T1 = P2V2/T2, the original pressure can be obtained as follows:

P1V1/T1 = P2V2/T2

P1 x 28/315 = 4 x 49/300

Cross multiply to express in linear form

P1 x 28 x 300 = 315 x 4 x 49

Divide both side by 28 x 300

P1 = (315 x 4 x 49) /(28 x 300)

P1 = 7.35atm

Therefore, the original pressure is 7.35atm

sure im new too!

explanation:

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.