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.
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.
the first question is question 4.
The second one is question 5
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.
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
To answer these questions, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Let's start with question 2:
1. First, we need to convert the initial temperature from Fahrenheit to Celsius. We have 68 °F, so we can use the formula °C = (°F - 32) * 5/9. Plugging in the values, we get:
°C = (68 - 32) * 5/9 = 36.67 °C.
2. Next, we need to convert the initial pressure from kPa to atm. We know that 1 atm = 101.325 kPa. So we can divide the given pressure of 106 kPa by 101.325 to get the pressure in atm: 106 kPa / 101.325 kPa/atm ≈ 1.045 atm.
3. Now, let's convert the initial volume from liters to milliliters and cups. We know that 1 L = 1000 mL and 1 L ≈ 4.227 cups. So we have:
Initial volume = 2.5 L = 2.5 * 1000 mL = 2500 mL
Initial volume ≈ 2.5 * 4.227 cups ≈ 10.567 cups.
4. To determine whether the bag will expand or contract, we need to compare the initial and final temperatures in Kelvin. The initial temperature is 36.67 °C, so we convert it to Kelvin by adding 273.15:
Initial temperature = 36.67 °C + 273.15 = 309.82 K.
The final temperature is -15 °C, so we convert it to Kelvin in the same way:
Final temperature = -15 °C + 273.15 = 258.15 K.
Since the final temperature is lower than the initial temperature, the gas will contract.
5. To find the new internal volume of the bag, we can use the combined gas law. The combined gas law can be derived from the ideal gas law by holding the number of moles constant. The formula is P₁V₁/T₁ = P₂V₂/T₂, where the subscripts 1 and 2 represent the initial and final states, respectively. Since the pressure is constant in this question, we can simplify the equation to V₁/T₁ = V₂/T₂.
Plugging in the values, we have:
V₁ = 2500 mL (initial volume)
T₁ = 309.82 K (initial temperature)
T₂ = 258.15 K (final temperature, in Kelvin)
We can solve for V₂ (new volume) by rearranging the equation:
V₂ = (V₁ * T₂) / T₁.
Plugging in the values, we get:
V₂ = (2500 mL * 258.15 K) / 309.82 K.
Performing the calculation:
V₂ ≈ 2084 mL.
To convert the volume back to liters and cups:
V₂ ≈ 2084 mL / 1000 mL/L ≈ 2.084 L
V₂ ≈ 2.084 L * 4.227 cups/L ≈ 8.798 cups.
So, the new internal volume of the bag is approximately 2084 mL, 2.084 L, or 8.798 cups.
Now let's move on to question 4:
1. We will use the same formula, V₁/T₁ = V₂/T₂, to find the final volume.
2. We already calculated the initial volume in question 2, which is 2.5 L or 2500 mL.
3. We are given the final temperature, which is 72 °F. We can convert it to Celsius using the formula °C = (°F - 32) * 5/9. Plugging in the values, we get:
°C = (72 - 32) * 5/9 = 40 °C.
4. Next, we convert the final temperature in Celsius to Kelvin:
Final temperature = 40 °C + 273.15 = 313.15 K.
5. We are asked to find the volume when the temperature equilibrates with room temperature. Since the volume doubles, the final volume will be double the initial volume.
Initially, the volume was 2.5 L or 2500 mL, so the final volume will be 2 * 2.5 L = 5 L or 2 * 2500 mL = 5000 mL.
To convert the final volume to cups, we have:
Final volume ≈ 5 L * 4.227 cups/L ≈ 21.135 cups.
So, the volume of the gas in the bag when the temperature equilibrates with room temperature is approximately 5000 mL, 5 L, or 21.135 cups.
I hope this explanation helps! Let me know if you have any further questions.