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A sample of oxygen gas occupies a volume of 0.4652 L. When the volume is increased to 0.5782 L, the new pressure is 341.3 atm. What was the original pressure? Gas Law used :_______________
n= pV/RT = 341.3 x 0.5782/8.314 x 300 = 0,079119421.p=nRT/V= 0,079119421 x 8.314 x 300/0.4652 = 424,203912 atm.
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An ideal gas starts with a pressure of 0.980 atm and a temperature of 23oC. The pressure raises to 1.1 atm. What is the new temperature of the gas?
P1/T1 = P2/T2T2 = P2 x T1 / p1 = 1.1atm x 300k / 0.980atm = 336.7k = 63.7oC
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At constant volume, a gas has a pressure of 120.0 kPa and a temperature of 30°C. The temperature is then increased to 191°C with no change in volume. What is the new pressure of the gas in kPa.
144.16kPa
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A sample of a gas occupies 250. mL at 1.00 atm of pressure. If the pressure increases to 2.00 atm while the temperature stays the same, what is the new volume?
P1V1 = P2V2P1 = 1V1 = 250 mlP2 = 2V2= ? 1×250 = 2V2 = 250/2V2 = 125 ml
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A 3.5 L sample at 360C and 5.2 atm contains 0.7 moles of a gas. If 0.3 mole of the gas is added to the sample without changing the temperature and pressure, what is the new volume?
If 0.3 mole is added, then total mole = 0.3+0.7=1PV=nRTV=nRT/P= 1 × 0.082 × 633 / 5.2=9.98L
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1. A confined amount of gas at 100C has a pressure of 2.5 atm. What is its new pressure if the temperature is increased to 50oC?
P1/T1 = P2/T22.5/373 = P2/3230.009156 = P2/323P2 = 0.009156×323P2 = 2.96atm
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A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 °C. What would its temperature be (in °C) when its pressure is increased to 11.2 atm?
For answering this question you need to use the next law: PV = nRT V, R, n are constants P is pressure T is temperature From the first law: P1/T1...
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A 12.0-L sample of a gas within a cylinder has a pressure of 2.0 atm. Determine the new pressure (in atm) when the volume of the cylinder decreases to 8.0-L.
p1V1 = p2V22.0 x 12 = p2 x 8p2 = (2.0 x 12) / 8 = 3 atm
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A sample of a gas occupies a volume of 1100 mL at 78ºC and 0.060 atm. What would be the new volume if the temperature decreased to 44ºC and the pressure increased to 1.31 atm?
p1 = 0,06 atm = 6079,5Pa V1 = 1100ml = 0,0011m3 T1 = 780C = 351 K p2 = 1,31 atm = 132735,8Pa T2 = 440C = 317 K V2...
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