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Lisa [10]
3 years ago
5

A sample of argon gas at 55°C is under 845 mm Hg pressure. What will the new temperature be if the pressure is raised to 1050 mm

Hg
Chemistry
1 answer:
Maurinko [17]3 years ago
7 0

Answer:

The final temperature at 1050 mmHg is 134.57 ^{\circ}C or 407.57 Kelvin.

Explanation:

Initial temperature = T = 55^{\circ}C = 328 K

Initial pressure = P = 845 mmHg

Assuming final  to be temperature to be T' Kelvin

Final Pressure = P' = 1050 mmHg  

The final temperature is obtained by following relation at constant volume

\displaystyle \frac{P}{P'}=\displaystyle \frac{T}{T'} \\ \displaystyle \frac{845 \textrm{ mmHg}}{1050 \textrm{ mmHg}} = \displaystyle \frac{328 \textrm{ K}}{T'} \\T' = 407.57 \textrm{ Kelvin}

The final temperature is 407.57 K

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____ KClO3-----&gt; ____ KCl + ____O3
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If an atom has an atomic number of 6 and and abdomens mass of 12 how many electrons does it have
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So, when atomic number is 6, proton number is also 6, and number of electrons will also be 6 in that atom.

Hope this helps! :)
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One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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3 years ago
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