1.A diver finds the best corals at a depth of approximately 10.0 m. The diver's lung capacity is 2.30 L. The air temperature is 30.42 °C and the pressure is 108.48 kPa. What is the volume of the diver's lungs at 10.00 m, at a temperature of 19.89 °C, and a pressure of 113.7 kPa?
Select one:
a. 2.1183
b. 1.435
c. 7.160
d. 1.771
The pressure of a 3.6 mL sample of gas at 97.3 ºC increases from 30.5 mm Hg to 49.5 mm Hg. If the volume is unchanged, what is the new temperature?
Select one:
a. 328
b. 601
c. -44.9
d. -71.4
The temperature of a sample of gas at 9.5 atm and 6.2 ºC increases to 91.9 ºC. If the volume is unchanged, what is the new gas pressure?
Select one:
a. 12.4
b. 141
c. 7.27
d. 93.0
59.88 liters of gas at atmospheric pressure is compressed into a 0.219 L gas cylinder. What is the pressure (mm Hg) of the compressed gas if its temperature remains constant?
Select one:
a. 2.08e5
b. 2.78
c. 58.0
d. 273
Chem help please!?
2016-08-03 5:19 am
回答 (1)
2016-08-03 5:45 am
1.
P₁ = 108.48 kPa, V₁ = 2.30 L, T₁ = (273.16 + 30.42) K = 303.58 K
P₂ = 113.7 kPa, V₂ = ? L, T₂ = (273.16 + 19.89) K = 293.05 K
For a fixed amount of gas :
P₁V₁/T₁ = P₂V₂/T₂
Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
New volume, V₂ = (2.30 L) × (108.48/113.7) × (293.05/303.58) L = 2.1183 L
...... The answer: a. 2.1183
====
2.
P₁ = 30.5 mmHg, T₁ = (273.2 + 97.3) K = 370.5 K
P₂ = 49.5 mmHg, T₂ = ? K
For a fixed amount of gas in constant volume :
P₁/T₁ = P₂/T₂
Then, T₂ = T₁ × (P₂/P₁)
New temperature, T₂ = (370.5 K) × (49.5/30.5) = 601 K = 328°C
...... If temperature is in K, the answer is: b. 601
...... If temperature is in °C, the answer is: a. 328
====
3.
P₁ = 9.5 atm, T₁ = (273.2 + 6.2) K = 279.4 K
P₂ = ? atm, T₂ = (273.2 + 91.9) K = 365.1 K
For a fixed amount of gas in constant volume :
P₁/T₁ = P₂/T₂
Then, P₂ = P₁ × (T₂/T₁)
New pressure, P₂ = (9.5 atm) × (365.1/279.4) = 12.4 atm
...... The answer: a. 12.4
====
4.
P₁ = 760 mmHg, V₁ = 59.88 L
P₂ = ? mmHg, V₂ = 0.219 L
For a fixed amount of gas under constant temperature :
P₁V₁ = P₂V₂
Then, P₂ = P₁ × (V₁/V₂)
New pressure, P₂ = (760 mmHg) × (59.88/0.219) = 2.08 × 10⁵ mmHg
...... The answer: a. 2.08 × 10⁵
P₁ = 108.48 kPa, V₁ = 2.30 L, T₁ = (273.16 + 30.42) K = 303.58 K
P₂ = 113.7 kPa, V₂ = ? L, T₂ = (273.16 + 19.89) K = 293.05 K
For a fixed amount of gas :
P₁V₁/T₁ = P₂V₂/T₂
Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
New volume, V₂ = (2.30 L) × (108.48/113.7) × (293.05/303.58) L = 2.1183 L
...... The answer: a. 2.1183
====
2.
P₁ = 30.5 mmHg, T₁ = (273.2 + 97.3) K = 370.5 K
P₂ = 49.5 mmHg, T₂ = ? K
For a fixed amount of gas in constant volume :
P₁/T₁ = P₂/T₂
Then, T₂ = T₁ × (P₂/P₁)
New temperature, T₂ = (370.5 K) × (49.5/30.5) = 601 K = 328°C
...... If temperature is in K, the answer is: b. 601
...... If temperature is in °C, the answer is: a. 328
====
3.
P₁ = 9.5 atm, T₁ = (273.2 + 6.2) K = 279.4 K
P₂ = ? atm, T₂ = (273.2 + 91.9) K = 365.1 K
For a fixed amount of gas in constant volume :
P₁/T₁ = P₂/T₂
Then, P₂ = P₁ × (T₂/T₁)
New pressure, P₂ = (9.5 atm) × (365.1/279.4) = 12.4 atm
...... The answer: a. 12.4
====
4.
P₁ = 760 mmHg, V₁ = 59.88 L
P₂ = ? mmHg, V₂ = 0.219 L
For a fixed amount of gas under constant temperature :
P₁V₁ = P₂V₂
Then, P₂ = P₁ × (V₁/V₂)
New pressure, P₂ = (760 mmHg) × (59.88/0.219) = 2.08 × 10⁵ mmHg
...... The answer: a. 2.08 × 10⁵
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