A graduated tube, sealed at the upper end, has a mercury leveling bulb connected

Question

A graduated tube, sealed at the upper end, has a mercury leveling bulb connected to the lower end by a flexible rubber tubing. The gas volume is 25.0 mL when the mercury level is the same in both tubes. The barometric pressure is 732 mm Hg. What is the gas volume when the mercury level on the open side is 10.0 cm above the mercury level the closed side? A sample of a gas at 0 degree C, and 1 atm pressure occupies 2.50 L What temperature change is necessary to adjust the pressure of the gas to 1.50 atmosphere when the gas has been transferred to a 2.00 L container? What volume of oxygen at 25 degree C and 725 mm Hg is needed to burn 2.00 L of butane, C_4H_10(g), at 725 mm Hg and 25 degree C? What volume of carbon dioxide would be produced at 350 degree C and 1.00 atm from the complete thermal decomposition of 1.00 kg of magnesium carbonate, MgCO_3? What is the density of carbon dioxide at this temperature? which contains only carbon, hydrogen, and sulfur is burned in oxygen under such t the individual volumes of the reactants and products can be measured at the same pressure. It is found that 3 volumes of the compound react with oxygen to yield 3

Explanation / Answer

we know that at STP(standard temperature and pressure) = 1 atm pressure and 273 K or 0 oC temperature
22.414 L of gas = 1 mole
1L = 1/22.414 mole
2.5 L = 1/22.414 * 2.5 = 0.1115 mole
here no. of moles remains constant as the same gas is transferred to less volume container

1 atm = 101325 Pa
new pressure 1.5 atm = 101325 * 1.5 Pa = 151987.5 Pa
1 L = 10-3 m3
new volume 2 L = 2 * 10-3 m3
R = 8.314 Pa * m3 /mol * K
0 oC = 273.15 K
new temperature = T

putting this in ideal gas equation
PV = nRT
151987.5 Pa * 2 * 10-3 m3 = 0.1115 mol * 8.314 Pa * m3 /mol * K * T

T = 327.9 K
change = 327.9 K - 273.15 K = 54.75 K change is needed

this can also be solved putting initial values to PV =nRT equation and then find out n
which will be same as above

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