Physics DSE Gas MC

2014-04-01 5:57 am

Two metallic containers X and Y of volume V and 4V respectively are connected by a narrow tube. Initially the tap S is closed and there is an ideal gas contained in X at pressure of 400 kPa and temperature 200 degree Celsius, while there is an ideal gas contained in Y at pressure of 100 kPa and temperature 400 degree Celsius.

If the tap S is then opened and assume the temperature of the container X and container Y remain constant at 200 degree Celsius and 400 degree Celsius respectively, what is the pressure of the ideal gas at equilibrium state?

Ans: 180kPa

Thanks!!

回答 (1)

天同

2014-04-01 6:57 am

✔ 最佳答案

Use ideal gas equation: PV = nRT
Container X: (400 x 10^3)V = (n1)R(273+200)
i.e. n1 = (400 x 10^3)V/473R
where n1 is the no. of moles of gas in container X

Container Y: (100 x 10^3)(4V) = (n2)R(273+400)
n2 = (4 x 10^5)V/(673R
where n2 is the no. of moles of gas in container Y

Hence, total no. of moles = n1 + n2 = [(4x10^5)V/R].(1/473 + 1/673)
= 1440(V/R)

After the tap is opened, let n1' and n2' be the new no. of moles of gas in the two containers X and Y respectively.
n1' = PV/473R and n2' = P(4V)/673R
(n1' + n2') = PV/473R + 4PV/673R
where P is the common pressure

Therefore, by conservation of mass,
n1 + n2 = n1' + n2'
1440(V/R) = (PV/R).(1/473+4/673)
P = 1440/(1/473 + 4/673) Pa = 179 000 Pa = 179 kPa

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