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ivanzaharov [21]
2 years ago
8

A gas initially at 273k is heated such that it volume and pressure became twice their original volume what is the new temperatur

e of the gas
Chemistry
1 answer:
lord [1]2 years ago
3 0

Answer:

<u>1092K</u>

Explanation:

We can use the combined gas law to answer this question:

P1V1/T1 = P2V2/T2,

where P, V and T are the Pressure, Volume, and Temperature for initial (1) and Final (2) conditions.  Temperatures must be in Kelvin.

The problem states that V2 = 2V1 and P2 = 2P1.

Let's rearrange to solve for T2, which is the question:

      T2 = T1(P2/P1)(V2/V1)

Note how the pressure and temperature values are written:  as ratios.  Enter the values:

       T2 = (273K)(P2/P1)(V2/V1)

        T2 = (273K)(2P1/P1)(2V1/V1)  [Use the expressions for V2 and P2 from above]

        T2 = (273K)(2)(2)

                       T2 = 1092K

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As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures lis
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Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
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when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
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by comparing the Kp by the Kp at equilibrium(the given value) So,
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∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.


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