Question

Submit A glass container was initially charged with 2.00 moles of a gas sample at 3.75 atm and 21.7 °C Some of the gas was released as the temperature was increased to 28.1 °C, so the final pressure in the container was reduced to 0.998 atm. How many moles of the gas sample are present at the end? mol 1 2 3 4 5 6 7 8 9 0 x 10

Answer 1

`Hey,

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1)

initial Moles = n_1 = 2.00 Moles initial Pressure = P_1 = 3.75 atm initial Temperature = T_1 = 21.7 degree C = (21.7 + 273) k T_1 = 294.7 k Final Temperature = 28.1 degree c = T_2 = (28.1 + 273) k T_2 = 301.1 k Final Pressure = P_2 = 0.998 atm Final Moles = n_2 =? ideal gas equation rightwardsdoublearrow PV = nRT P_1 V_1/T_1 n_1 = R rightwardsdoublearrow P_1/T_1 n_1 = R/V = constant V = constant R =constant P_1/T_1 n_1 = P_2/T_2 n_2 P_1 T_2 n_2 = P_2 T_1 n_1 n_2 = P_2 T_1 n_1/P_1 T_2 = 0.998 times 294.7 times 2/3.75 times 301.1 = 588.22/1129.125 n_2 = 0.52 mole Final Moles of gas = [0.52 Mole]\r\n Ideal gas equation is PV = nRT Where T = Temperature = 75.3 degree C = 75.3 +273 = 348.3 K P = pressure = 3.00 atm n = No. of moles = 20.0 mol

2)

2. Ideal gas equation is PV = nRT

Where

T = Temperature = 75.3 oC = 75.3 +273 = 348.3 K

P = pressure = 3.00 atm

n = No . of moles = 20.0 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = ?

Plug the values we get V = (nRT)/P = 190.6 L

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