Question

2. A gas syringe at 20°C contains 20.0 mL of CO2 gas. The pressure of the gas in the syringe is 1.0 atm. What is the pressure in the syringe at 20°C if the plunger is depressed to 10.0 mL?



3. A sample of SO2 gas has a volume of 5.2 L. It is heated at constant pressure from 30.o oC to 90.0°C. What is its new volume? (Hint: Which law will you apply of the pressure is constant

4. A sample of the inhalation anesthetic gas Halothane, in a 500-mL cylinder has a pressure of 2.3 atm at 0°C. What will be the pressure of the gas if its temperature is warmed to 37°C (body temperature)? (Hint: Which law will you apply if the volume is constant?)

Answer 1

Answer: 2. 2.0 atm

3. 6.2 L

4. 2.6 atm

Explanation:

2. To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=1.0atm\\V_1=20.0mL\\P_2=?\\V_2=10.0mL$

Putting values in above equation, we get:

$1.0\times 20.0mL=P_2\times 10.0mL\\\\P_2=2.0atm$

Thus new pressure in the syringe at 20°C if the plunger is depressed to 10.0 mL is 2.0 atm

3. To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=5.2L\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=90.0^oC=(90.0+273)K=363K$

Putting values in above equation, we get:

$\frac{5.2L}{303K}=\frac{V_2}{363}\\\\V_2=6.2L$

Thus  its new volume is 6.2 L.

4. To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=2.3atm\\T_1=0^0C=273K\\P_2=?\\T_2=37^0C=(37+273)K=310K$

Putting values in above equation, we get:

$\frac{2.3atm}{273K}=\frac{P_2}{310K}\\\\P_2=2.6atm$

Hence, the pressure of the gas if its temperature is warmed to 37°C is 2.6 atm

Answer 2

Answer:

2. Based on the Boyle's law, the pressure of the sample of a gas is inversely proportional to the volume at a constant temperature. Mathematically, the law can be written as,

P₁V₁ = P₂V₂

Here P₁ and V₁ are the pressure and volume initially, while P₂ and V₂ are the eventual pressure and volume.

Based on the given information, P₁ is atm, V₁ is 20 ml, V₂ is 10 ml and there is a need to find P₂. The volumes V₁ and V₂ can also be written as,

V₁ = 20 ml × 1L / 1000 ml = 0.02 L

V₂ = 10 ml × 1L / 1000 ml = 0.01 L

Now putting the values, we get P₂ as,

P₂ = P₁V₁/V₂

P₂ = 1 atm × 0.02 L / 0.01 L

P₂ = 2 atm

3) Based on the given question, the initial volume (V₁) of the sample is 5.2 L, the initial temperature (T₁) is 30 degree C or 273 + 30 = 303 K, the final temperature (T₂) is 90 degree C or 273 + 90 = 363 K.

Based on Charle's law, the volume of the gas is directly proportional to the temperature at constant pressure,

V = KT, here K is the proportionality constant

K or constant = V/T

Now using the formula,

V₁/T₁ = V₂/T₂

5.2/303 = V₂/363

V₂ = (5.2 × 363) / 303

V₂ = 6.22 L

4. Based on the given information, the T₁ or initial temperature of halothane is 0 degree C or 273 + 0 = 273 K, the T₂ or the final temperature of halothane is 37 degree C or 273 + 37= 310 K.

The P₁ or initial pressure of halothane is 2.3 atm, and as volume is constant so V₁ = V₂

Now using the combined gas law equation, we can find P₂,

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁T₂/T₁V₂

P₂ = (2.3 atm) (310 K) / (273 K)

P₂ = 2.6 atm

Hence, the final pressure of halothane is 2.6 atm.