Problem 1 A flask of volume o.s liters contains o.022 molen of an ideal gas conn

Question

Problem 1 A flask of volume o.s liters contains o.022 molen of an ideal gas connected (101 kpa) and a temperature of 273 K. The flask is to the atmosphere through a topcock, which is open. The flask immersed in a bath of water boiling at the steam allowed to point and come to thermal equilibrium. The is then closed and the flask is transferred to an ice and allowed to come to equilibrium. bath 1. When the flask is in the bath of boiling water, the pressure in the flask is kPa 101 B. 138 C. 74 276 2 When the flask has come to equilibrium in the icebath with its stopcock closed, the pressure in the flask is kPa. 101 E- 276 138 37 3. The flask contains moles of gas when in equilibrium at the steam point A. 0.022 B. 0.016 0.008 0.030 The flask contains moles of gas when in equilibrium at the ice point with its stopcock closed. 0.008 E. 0.060 A. 0.022 B. 0.016 0.030 blem 2 has a hole with a 25.000 om circumference at a temperature of 20 The coefficient of linear expansion for aluminum is 24x10 .c 1. At o c the circumference of the hole is cm. E. 25.000 C. 25.024 D. 24.976 B. 24.988 A. 25.012 The circumference of the hole will be 25.100 cm at a temperature of 187 147 100

Explanation / Answer

1)
p1 = 101 Kpa

V1 = 0.5 l

n = 0.022

T = 273 K

a)

in boiling water , as Pressure is connected to atmosphere ,

P = 101 Kpa

b)

in the boiling water , let n2 be moles of gas

then, using ideal gas

n1*T1 = n2*T2

273 *0.022 = n2*373

n2 = 0.016 moles

Now ,

Using ideal gas equation

P = .016*8.314*273/.5*10^-3

P = 74 Kpa

Pressure is 74 Kpa

3)

flask contain 0.016 moles as calculated above

4)

as the stopcock is closed ,

n is fixed

and moles of gas is 0.016

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