Solve the following problems and show your complete solution. (USE GUESA)
1. An argon gas occupies 56.2 L at 760 torr. If the volume of gas is decreased to 1⁄4 of the original volume, calculate its
final pressure.
2. At 0OC and 5 atm, a given sample of gas occupies 75 L. The gas is compressed to a final volume of 30 L at 0OC. What
is the final pressure?
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1. An argon gas occupies 56.2 L at 760 torr. If the volume of gas is decreased to 1⁄4 of the original volume, calculate its
final pressure.
2. At 0OC and 5 atm, a given sample of gas occupies 75 L. The gas is compressed to a final volume of 30 L at 0OC. What
is the final pressure?
brainliest ko ang Tama
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Answer:
1. Using the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to find the number of moles of argon gas present in the original state:
PV = nRT
n = PV/RT
n = (760 torr)(56.2 L) / (0.0821 L·atm/mol·K)(273 K)
n = 2.19 mol
Next, we can use Boyle's law to determine the final pressure when the volume is decreased to 1/4 of the original volume:
P1V1 = P2V2
(760 torr)(56.2 L) = P2(56.2 L/4)
P2 = (760 torr)(56.2 L) / (56.2 L/4)
P2 = 3040 torr
Therefore, the final pressure of the argon gas is 3040 torr.
2. Since the temperature is constant at 0°C, we can use the combined gas law equation: P1V1/T1 = P2V2/T2, where P is pressure, V is volume, and T is temperature in Kelvin.
We need to convert the pressure to the appropriate units of measure, so we use the conversion factor 1 atm = 101.325 kPa:
P1 = 5 atm × 101.325 kPa/atm = 506.625 kPa
Next, we can substitute the given values into the combined gas law equation:
P1V1/T1 = P2V2/T2
(506.625 kPa)(75 L)/(273 K) = P2(30 L)/(273 K)
P2 = (506.625 kPa)(75 L)/(30 L)(273 K)
P2 = 4.33 atm
Therefore, the final pressure of the gas is 4.33 atm.