2. A stretchy rubber bag is filled with 2.5 L nitrogen gas at a temperature of 68 °F under a constant pressure of 106 kPa. The bag is placed in the freezer where the temperature is −15 °C. Does the bag expand or contract? What is the new internal volume of the bag? Express your answer in liters, milliliters, and cups.

Question

4. The stretched bag in question 4 has more nitrogen gas added while in the freezer until the volume doubles. The bag is returned to the room environment where the temperature has risen to 72 °F. What is the volume of the gas in the bag when the temperature of the nitrogen equilibrates with room temperature? Express your answer in liters, milliliters, and cups.

Answer 1

the first question is question 4.
The second one is question 5

Answer 2

2.
Use V1/T1 = V2/T2
V1 is 2.5 L
V2 = ?
T1 = 68 F = 20 C = ? K.............K = 273 + 20 = ?
T2 = 15 C. T2 = 273 + 15 = K
Substitute the values (use K for T1 and T2)

4 is worked the same way but you omitted the units of how much N2 was added to the bag before taking it out of the fridge.

Post your work if you get stuck.

Answer 3

Question 4
P V = n R T
Psame , n same, R same
so
V = constant * T
if T goes down, V goes down (but we all knew it shrank)
we need T in Kelvin, Centigrade + 273
Kelvin T at start =(5/9) * (F + 460) =(5/9)(68+460) = 293deg K
Kelvin T at end = C+273= 273 - 15 = 258 deg K
V/T = 2.5 L / 293 = newV/258
new V = (258/293) 2.5 liters = 2.20 liters, You do mL and silly cups

Question 5
add gas until V doubles
P V = n R T
If it is real stretchy P R and T do not change
so
if V doubles, n doubles
now let it warm up
It will still occupy twice the volume at 293
but now it warms up to (5/9)(72+460) = 296deg K
V = (296/293) * 2 * 2.5 L = 5.05 Liters