You place a balloon in a closed chamber at standard temperature and pressure (STP) . You double the chamber temperature while leaving the pressure constant. What happens to the balloon?

Volume of balloon will be doubled

Explanation:

We are asked to state what happens to a balloon which was placed in a closed chamber at STP if the temperature is doubled and the pressure is kept constant.

We have three parameters here, the temperature and pressure of the chamber, and the volume of the balloon.

Now, since the pressure of the chamber is kept constant, we can say that the gas law we need to use is the Charles’ law and it posits that volume and temperature are directly proportional:

Given;

V1 =? , V2 = ?, T1 = 273.15K(at STP) and T2 = 273.15 * 2 = 546.3K

We substitute these values in the Charles’ law equation:

V1/T1 = V2/T2

V2 = V1T2/T1

V2 = (V1 * 546.3)/273.15

V2 = 2V1

This means that the balloon will double in volume

yes, water is wet, just as it makes other things wet, it is wet itself, kinda like ice, when ice is in water, the ice makes water cold, and ice is cold, therefore with that example, water is wet

i believe it is c.) 1575 j.