Chapter 5
Q. 5.2
A sealed 15.0-L steel tank is used to deliver propane (C_3H_8) gas. It is filled with 24.6 g of propane at 27°C. The pressure gauge registers 0.915 atm. (Assume that the expansion of steel from an increase in temperature is negligible.)
ⓐ If the tank is heated to 58°C, what is the pressure of propane in the tank?
ⓑ The tank is fitted with a valve to open and release propane to maintain the pressure at 1.200 atm. Will heating the tank to 58°C release propane?
ⓒ At 200°C, the pressure exceeds 1.200 atm. How much propane is released to maintain 1.200 atm pressure?
ⓐ
ANALYSIS | |
V (15.0 L); P (0.915 atm); T (27°C); mass of propane (24.6 g); T (27°C); T (58°C) |
Information given: |
2 sets of conditions for temperature | Information implied: |
pressure after the temperature is increased | Asked for: |
STRATEGY
1. Given two sets of conditions, you need to use the formula for initial state-final state conditions.
2. A sealed steel tank implies that the number of moles and the volume are kept constant.
3. Make sure all temperatures are in K.
ⓑ
ANALYSIS | |
from part (a): P (1.01 atm); T (58°C); condition for valve to open (1.200 atm pressure) |
Information given: |
Will the valve open? | Asked for: |
ⓒ
ANALYSIS | |
V (15.0 L); P (0.915 atm); T (27°C); mass of propane (24.6 g) P (1.200 atm); T (200°C) |
Information given: |
2 sets of conditions for temperature and pressure | Information implied: |
mass of propane released | Asked for: |
STRATEGY
1. Convert grams of propane to moles and temperatures in °C to K.
2. Given two sets of conditions, you need to use the formula for initial state-final state conditions to find the number of moles of propane related to the second set of conditions.
3. The steel tank implies that the volume is kept constant.
Step-by-Step
Verified Solution
ⓐ
P_{2} | \frac{V_{1} P_{1}}{n_{1} T_{1}} = \frac{V_{2} P_{2}}{n_{2} T_{2}} → \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} → \frac{0.915}{27 + 273} = \frac{P_{2}}{58 + 273} → P_{2} = 1.01 atm |
ⓑ
Will the valve open? | Valve opens at 1.200 atm. 1.01 (from part (a)) < 1.200 The valve will not open. |
ⓒ
c) mol C_{3}H_{8} initially (n_{1}) | 24.6 g × \frac{1 mol}{44.1 g} = 0.558 mol |
Initial conditions | P_{1} = 0.915 atm; T_{1} = 27°C + 273 = 300 K; n_{1} = 0.558 mol; V_{1} = 15.0 L |
Final conditions | P_{2} = 1.200 atm; T_{2} = 200°C + 273 = 473 K; n_{2} = ?; V_{2} = 15.0 L |
n_{2} | \frac{V_{1} P_{1}}{n_{1}T_{1}} = \frac{V_{2} P_{2}}{n_{2} T_{2}} → \frac{0.915}{0.558 × 300} = \frac{1.200}{n_{2} × 473},n_{2} = 0.464 mol |
Mass C_{3}H_{8} in the tank | (0.464 mol)(44.1 g/mol) = 20.5 g |
Mass to be released | 24.6 − 20.5 = 4.1 g |
END POINT
Note that the volume of the tank is never used in the calculations. It is important not only to read the problem carefully but also to visualize the description of the gas container. If the gas were in a balloon, instead of in a steel tank, the calculations would be different.