What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 33 ∘C , V1= 1.36×103 L ) to a volume of 2.72×103 L ?

Question

Assume no change in pressure or the amount of gas in the balloon.

Answer 1

For this problem, we use the Combined Ideal Gas Law, simplified to include just temperature and volume, since pressure and moles stay the same. This is Charles' Law: V1/T1 = V2/T2. We're asked for T2, so we solve by multiplying both sides by T2, dividing by V1, and multiplying by T1 to get

T2 = (V2*T1)/V1.

What's really important to remember is that temperatures must always be positive (on an absolute scale) to get solutions that make sense: we need to have the temperatures in K for all ideal gas problems. So we have T1 = 33 + 273 = 306 K; we're given V1 = 1.36 x 10^3 L and V2 = 2.72 x 10^3 L. Filling in to the equation:

T2 = (2.72 x 10^3 L * 306 K)/1.36 x 10^3 L = 612 K.

Temperature and volume are directly proportional: if you increase one, the other will increase. (Think of heating a hot-air balloon to make it rise, or taking a helium balloon from a warm house outside on a cold winter day). So it makes sense that you would need a higher temperature to increase the volume. In fact, V2 is exactly double V1, so T2 is double T1.

Answer 2

Simply put, you are asking what temperature should a gas be heated to in order to double its volume..Temoerature one is 33^OC, and starting volume is 1.36 x 10³L; final volume is 2.72 x 10³L, which is exactly double the initial volume. therefore, the temperature,T₁, should be multiplied by the ratio of temperatures:

T2 = T1 x 2.72/1.36 = 33^OC x 2 = 66^OC