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The mass of a jar of beans is 525 g. If the jar has a mass of 25 g, what is the mass of the beans alone?
If the total mass consists of the mass of beams and the mass of the jar: m(total) = m(beams) + m(jar) Therefore, the mass of beams alone equals: m(beams) =...
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A 2.00 L sample of gas originally at 25 °C and pressure of 400 torr is allowed to expand
to a volume of 5.00 L.
If the final pressure of gas is 585 torr, what is the final temperature in K?
p*V = n*R*Tn and R are constant so p*V/T = constantThen (2.00 L)*(400 torr)/(25+273.15 K) = (5.00 L)*(585 torr)/TT = (5.00 L)*(585 torr)*(25+273.15 K)/(2.00 L)/(400 torr) = 1090 K
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Four moles of an ideal gas expand isothermally from 1litre to 10 litre at 300kelvin.calculate the change in free energy of the gas.(universal gas constant=8.314)
Change of free energy= nRT lnV1/V2= 4*8.314*300ln (1/10=-22972 J
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If the total air pressure is 1.05 atm, the partial pressure of carbon dioxide is 0.08 atm, and the partial pressure of hydrogen sulfide is 0.05 atm, what is the partial pressure of the remaining air?
As we know, Total pressure is the sum of partial pressure of individual gases.PT = 1.05 -(0.08+0.05+Pair)Pair = [0.92] atm
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A gas with a pressure of 557 mmHg and a volume of 634 mL is allowed to expand to a volume of 975 mL. What is the resulting pressure in atm?
Solution:Convert mmHg to atm:(557 mmHg) × (1 atm / 760 mmHg) = 0.733 atmP1 = 557 mmHg = 0.733 atmV1 = 634 mLT1 = T2 = constP2 = unknownV2 =...
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Container A holds 782 mL of ideal gas at 2.80 atm. Container B holds 174 mL of ideal gas at 4.40 atm. If the gases are allowed to mix together, what is the resulting pressure?
Answer on Question #83513Container A holds 782 mL of ideal gas at 2.80 atm. Container B holds 174 mL of ideal gas at 4.40 atm. If the gases are allowed...
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A gas occupying a volume of 625cm3 at a pressure of 90KPA is allowed expand at constant temperature till reduced to 50 KPA. What is the final volume of the gas?
1125 cm3
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If some neon gas at 121,000 Pa were allowed to expand from 37 mL to 60 mL without changing the temperature, what pressure would the neon gas exert under these new conditions?
pV=nRT. n=0.121 x 37 / 8.314 x 273. = 0,00197248826.p= nRT/V=0,00197248826 x 8.314 x 273 / 60 = 0,0746166667 kpa.74.616 pa.
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One mole of a perfect gas at 300K has an initial pressure at 15 atm and is allowed to contract isothermally to a pressure of 1 atm. Calculate the entropy change from this contraction
Initial pressure=15atm,final pressure=1atmTemperature=300kNow 1Latm= 103joules∆dsystem=∆dsurrounding=-(dv)/T={-(15-1)/300}×103=-4.80J.
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In a gas mixture composed of N2, Ne, and He, the partial pressure of N2 is 0.50 atm, that of Ne is 1.1 atm, and that of He is 0.80 atm. Calculate the mole fraction of each gas.
Mole fraction of N2- 0.208Mole fraction of Ne - 0.3Mole fraction of He - 0.458
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