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Consider two bulbs seperated by a valce. Both bulbs are amintained at the same temperature. Assume that when the valve between the two bulbs is closed, the gases are sealed in their respective bulbs. When the valve is closed, the following data apply:

Bulb A Bulb B
Gas Ne CO
V 2.50L 2.00L
P 1.09 atm 0.73 atm


Assuming no temperature change, determine the final pressure inside the system after the valve connecting the two bulbs is opened. Ignore the volume of the tube connecting the two bulbs.

User Chink
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1 Answer

4 votes

Answer:

The pressure is
P_f = 0.93 \ atm

Step-by-step explanation:

From the question we are told that

The volume of Ne is
V_N = 2.50 \ L

The volume of CO is
V_C = 2.00 \ L

The pressure of
Ne is
P_N = 1.09 \ atm

The pressure of CO is
P_C = 0.773 \ atm

The number of moles of Ne present is evaluated using the ideal gas equation as


n_N = (P_N * V_N)/(R T)

=>
n_N = (1.09 * 2.50 )/(R T) = (2.725)/(RT)

The number of moles of CO present is evaluated using the ideal gas equation as


n_N = (P_C * V_C)/(R T)

=>
n_N = (0.73 * 2.00 )/(R T) = (1.46)/(RT)

The total number of moles of gas present is evaluated as


n_T = n_N + n_C


n_T = (2.725)/(RT) + (1.46)/(RT)


n_T = (4.185)/(RT)

The total volume of gas present when valve is opened is mathematically represented as


V_T = V_N + V_C

=>
V_T = 2.50 + 2.00 = 4.50 \ L

So

From the ideal gas equation the final pressure inside the system is mathematically represented as


P_f = (n_T * RT )/( V_T)

=>
P_f = ([(4.185)/(RT) ] * RT )/( 4.50)

=>
P_f = 0.93 \ atm

User Timothy Walters
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