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The pressure of a sample of helium in a 1.50-L container is 0.88 atm. What is the new pressure if the sample is placed in a 2.00 L container?
P1V1 = P2V20.88 atm × 1.50L = 2.00L × X1.32 = 2.00XX = 1.32/2.00= 0.66 atm
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20. If 4.9 g of nitrogen gas is placed in a 2 L container at 31 °C, what is the pressure in the container?
Pressure =2.1atm
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Three samples of gas each exert 740. mm Hg in separate 2 L containers. What pressure do they exert if they are all placed in a single 2 L container
Presure exerted =1480mmHg
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A gas exerts a pressure of 2.0 atm, at 30°C, in a 10 L container. In what size container would the same amount of gas exert a pressure of 4.0 atm at 20° C?
Using the gas law: pV = nRT, and keeping the same number of moles of gas (n),we find the final volume V2:V2 = (T2/p2)*(p1*V1/T1) = (293 K/ 4.0 atm)*(2.0 atm...
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0.740 moles of a gas occupies 20.0 L at 700.0 mm Hg. What will the volume be when four moles of gas are added to the container and the new pressure is 800.0 mm Hg. Assume no change in temperature
n1 = 0.740 molesn2 = 0.740 moles + 4.000 moles = 4.740 molesP1 = 700.0 mm HgV1 = 20.0 LP2 = 800.0 mm HgV2 = unknownT1 = T2 = T...
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When a real gas is placed into a container at high pressure, why might the pressure of the gas be higher than what was predicted by the ideal gas law?
The weight and size of the molecules that are negotiated in the ideal gas law increase the volume of the gas mixture. However, as the container has a constant volume,...
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A mixture of four gases exerts a total pressure of 860mmHg. Gas A and B each exert 220mmHg. Gas C exerts 110mmHg. What pressure is exerted by gas D?
For this task we use the Dalton’s law which says that total pressure of mixture of gases equals to the sum of partial pressures of individual gases: Ptotal=PA+PB+PC+PD Therefore PD...
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a sample oh hydrogen gas, H2(g), is placed in a 0.500L container at 295 K. The gas pressure is 1.442bar. how many moles of H2 gas are in the container
pV=nRTT = 295 KV = 0.5 L = 0.5*10-3 m3p = 1.442 bar = 144200 Pa R = 8.314 J/mol*Kn = pV/RT = (144200*0.5*10-3)/(8.314*295) = 0,029 mol.
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Consider a container with 9.05g of N2 gas and an unknown mass of CO2 gas. If we know that the partial pressure of the CO2 is twice that of the N2, determine the mass of CO2 gas in the container.
P is proportional to nPco2/PN2 = nco2/nN2Moles of gas is equal to the weight/molecular weight of gasPco2 is twice that of PN2Pco2 = 2PN2, weight of N2 = 9.05 gram2PN2/PN2...
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Consider a container with 5.08g of N2 gas and an unknown mass of CO2 gas. If we know that the partial pressure of the CO2 is twice that of the N2, determine the mass of CO2 gas in the container
Mass of nitrogen= 5.08 gNo.of mole of nitrogen= 5.08/28= .182Partial pressed of nitrogen let = PPV=nRTV=.182RT/PIn case of carbon dioxideMass of carbon dioxide=?Partial pressure of carbon dioxide= 2PPV=nRTn=PV/RTn=2P×.182RT/PRTn=.364 Mass of...
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