Question

An ideal gas is contained in two metal cylinders X and Y connected by a tap which is initially closed. The volume and pressure of the gas in the cylinders are as attached.When the tap connecting the two cylinders is opened, what will be the final pressure, in Pa, in the vessel? Assume that the temperature remains constant
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2.4×105

3.5×105

4.2×105

5.0×105

Answer 1

Answer on Question #51933-Physics-Field Theory

An ideal gas is contained in two metal cylinders X and Y connected by a tap which is initially closed. The volume and pressure of the gas in the cylinders are as attached.

When the tap connecting the two cylinders is opened, what will be the final pressure, in Pa, in the vessel? Assume that the temperature remains constant

2.4×105

3.5×105

4.2×105

5.0×105

Solution

We have

$$P _ {X} V _ {X} = \frac {m _ {X}}{M} R T, P _ {Y} V _ {Y} = \frac {m _ {Y}}{M} R T, P (V _ {X} + V _ {Y}) = \frac {(m _ {X} + m _ {Y})}{M} R T.$$

So,

$$P _ {X} V _ {X} + P _ {Y} V _ {Y} = P \left(V _ {X} + V _ {Y}\right).$$

Thus,

$$P = \frac {P _ {X} V _ {X} + P _ {Y} V _ {Y}}{V _ {X} + V _ {Y}} = \frac {5 \cdot 1 0 ^ {5} \cdot 1 1 \cdot 1 0 ^ {- 3} + 2 \cdot 1 0 ^ {5} \cdot 4 \cdot 1 0 ^ {- 3}}{(1 1 + 4) \cdot 1 0 ^ {- 3}} = 4. 2 \cdot 1 0 ^ {5} P a.$$

Answer: $4.2 \cdot 10^{5} Pa$ .

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