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What is the pressure of 4 moles of helium in a 50L tank at 308K Use PVnRT?
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∙ 7y ago
2.02 atm
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What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for pressure. Plugging in the values, the pressure of the 4 moles of helium in a 50 liter tank at 308 K is approximately 81.6 atm.
How many moles of gas are present in a 10L container when 3.84atm of pressure is exerted at 35 Celsius?
Using the ideal gas law (PV = nRT), we can calculate the number of moles of gas present in the container. First, convert the temperature to Kelvin (35°C + 273 = 308K). Then plug in the values: 3.84 atm * 10 L = n * 0.0821 Latm/molK * 308 K to solve for n. This gives approximately 1.5 moles of gas in the container.
Why co2 cannot be liquified at 308k?
At 308K, carbon dioxide is above its critical temperature of 304.25K. This means that it cannot exist as a liquid under those conditions, as it would be above its critical point and would behave as a supercritical fluid instead of distinct liquid and gas phases.
What is the pressure of 4 moles of helium in a 50 liter tank at 308k?
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for pressure. Plugging in the values, the pressure of the 4 moles of helium in a 50 liter tank at 308 K is approximately 81.6 atm.
How many moles of ammonia gas are found in a 202 mL container at 35C and 750 mmHg?
To find the number of moles of ammonia gas, you can use the ideal gas law equation: PV = nRT. Convert the volume to liters (202mL = 0.202L) and the temperature to Kelvin (35°C + 273 = 308K). Plug in the values: (0.750 atm) * (0.202 L) = n * (0.0821 Latm/molK) * (308K), solve for n to find the number of moles of ammonia gas.
How many moles of gas are present in a 10L container when 3.84atm of pressure is exerted at 35 Celsius?
Using the ideal gas law (PV = nRT), we can calculate the number of moles of gas present in the container. First, convert the temperature to Kelvin (35°C + 273 = 308K). Then plug in the values: 3.84 atm * 10 L = n * 0.0821 Latm/molK * 308 K to solve for n. This gives approximately 1.5 moles of gas in the container.
Why co2 cannot be liquified at 308k?
At 308K, carbon dioxide is above its critical temperature of 304.25K. This means that it cannot exist as a liquid under those conditions, as it would be above its critical point and would behave as a supercritical fluid instead of distinct liquid and gas phases.
10.0 grams of a gas occupies 12.5 liters at a pressure of 42.0 mm Hg What is the volume when the pressure has increased to 75.0 mm Hg?
Using Boyle's Law, p1*V1= p2*V2. This means that the pressure multiplied by the volume remains constant whilst the temperature is the same. Therefore; p1=42.0mm Hg, V1= 12.5L and so the product of the two is 525. If the pressure is now 75 mm Hg the volume must be 525/75= 7 liters. The 10.0 grams of gas information is not needed.
What is the volume in liters occupied by 1.21g of Freon-12 gas at 0.980 ATM and 308k?
V = m*R_sp*T/p = 1.21*(8.314/121)*308/(0.98*101325) = 2.58*10^-4 [m^3] = 0.258 [L]
