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Deaf_Jayme

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Answer the following problems:

1. Helium gas has a volume of 250 mL at 0°C at 1.0 atm. What will be the final pressure if the volume is reduced to 100 mL at 45°C?

2. The volume of a gas at 27°C and 700.0 mmHg is 600.0 mL. What is the volume of the gas at -20.0°C and 500.0 mmHg?

3. A 2.5 L of nitrogen gas exerts a pressure of 760 mmHg at 473 K. What temperature is needed to reduce the volume to 1.75 L at 1140 torr?
 
1. To solve this problem, we need to use the combined gas law formula: P1V1/T1 = P2V2/T2

P1 = 1.0 atm
V1 = 250 mL
T1 = 273 K (0°C + 273)
P2 = ?
V2 = 100 mL
T2 = 318 K (45°C + 273)

Plugging these values into the formula, we get:

1.0 atm * 250 mL / 273 K = P2 * 100 mL / 318 K

Simplifying this equation, we get:

P2 = 1.42 atm

Therefore, the final pressure will be 1.42 atm.

2. For this problem, we can again use the combined gas law formula:

P1V1/T1 = P2V2/T2

P1 = 700.0 mmHg
V1 = 600.0 mL
T1 = 300 K (27°C + 273)
P2 = 500.0 mmHg
V2 = ?
T2 = 253 K (-20.0°C + 273)

Plugging these values into the formula, we get:

700.0 mmHg * 600.0 mL / 300 K = 500.0 mmHg * V2 / 253 K

Simplifying this equation, we get:

V2 = 432.7 mL

Therefore, the volume of the gas at -20.0°C and 500.0 mmHg is 432.7 mL.

3. For this problem, we can use the ideal gas law formula: PV = nRT

We can first calculate the number of moles of nitrogen gas using the initial conditions:

P1 = 760 mmHg
V1 = 2.5 L
T1 = 473 K
R = 0.0821 L·atm/(mol·K)

PV = nRT
n = PV/RT
n = (760 mmHg * 2.5 L) / (0.0821 L·atm/(mol·K) * 473 K)
n = 0.211 mol

Now that we know the number of moles, we can use the ideal gas law again to solve for the final temperature:

P2 = 1140 torr
V2 = 1.75 L
n = 0.211 mol
R = 62.36 torr·L/(mol·K)

PV = nRT
T = PV/nR
T = (1140 torr * 1.75 L)/(0.211 mol * 62.36 torr·L/(mol·K))
T = 539 K

Therefore, the temperature needed to reduce the volume to 1.75 L at 1140 torr is 539 K.
 
Thank you so much for help me

1. To solve this problem, we need to use the combined gas law formula: P1V1/T1 = P2V2/T2

P1 = 1.0 atm
V1 = 250 mL
T1 = 273 K (0°C + 273)
P2 = ?
V2 = 100 mL
T2 = 318 K (45°C + 273)

Plugging these values into the formula, we get:

1.0 atm * 250 mL / 273 K = P2 * 100 mL / 318 K

Simplifying this equation, we get:

P2 = 1.42 atm

Therefore, the final pressure will be 1.42 atm.

2. For this problem, we can again use the combined gas law formula:

P1V1/T1 = P2V2/T2

P1 = 700.0 mmHg
V1 = 600.0 mL
T1 = 300 K (27°C + 273)
P2 = 500.0 mmHg
V2 = ?
T2 = 253 K (-20.0°C + 273)

Plugging these values into the formula, we get:

700.0 mmHg * 600.0 mL / 300 K = 500.0 mmHg * V2 / 253 K

Simplifying this equation, we get:

V2 = 432.7 mL

Therefore, the volume of the gas at -20.0°C and 500.0 mmHg is 432.7 mL.

3. For this problem, we can use the ideal gas law formula: PV = nRT

We can first calculate the number of moles of nitrogen gas using the initial conditions:

P1 = 760 mmHg
V1 = 2.5 L
T1 = 473 K
R = 0.0821 L·atm/(mol·K)

PV = nRT
n = PV/RT
n = (760 mmHg * 2.5 L) / (0.0821 L·atm/(mol·K) * 473 K)
n = 0.211 mol

Now that we know the number of moles, we can use the ideal gas law again to solve for the final temperature:

P2 = 1140 torr
V2 = 1.75 L
n = 0.211 mol
R = 62.36 torr·L/(mol·K)

PV = nRT
T = PV/nR
T = (1140 torr * 1.75 L)/(0.211 mol * 62.36 torr·L/(mol·K))
T = 539 K

Therefore, the temperature needed to reduce the volume to 1.75 L at 1140 torr is 539 K.
Thank you so much for the help me
 
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