Question #a651a

1 Answer
Oct 11, 2015

8.3^@"C"8.3C

Explanation:

Notice that no mention was made about the number of moles of gas and the volume of the cylinder, which means that you can ssume them to be constant.

In this case, you know that pressur eand temperature have a direct relationship when the number of moles of gas and the volume of the container are kept constant - this is known as Gay Lussac's Law.

In other words, if the temperature increases, the pressure increases as well. If the temperature decreases, the pressure decreases as well.

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In your case, the pressure of the sample of gas decreased from "4.88 atm"4.88 atm the first day, to "4.69 atm"4.69 atm the second day.

This means that the temperature of the sample decreased as well,so you can expect the temperature on th previous day to be a little higher than 8^@"C"8C.

Mathematically, Gay Lussac's Law can be written like this

(P_1)/T_1 = P_2/T_2" "P1T1=P2T2 , where

P_1P1, T_1T1 - the pressure and temperature of the gas at an initial state;
P_2P2, T_2T2 - the pressure and temperature of the gas at a final state.

Plug in your values and solve for T_1T1

T_1 = P_1/P_2 * T_2T1=P1P2T2

T_1 = (4.88color(red)(cancel(color(black)("atm"))))/(4.69color(red)(cancel(color(black)("atm")))) * 8^@"C" = 8.324^@"C"

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the temperature of the gas

T_1 = color(green)(8.3^@"C")