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Chemisorption

One mole of a...

Question

One mole of an ideal gas is compressed from $500 c m^{3}$ against a constant pressure of $1.216 \times 10^{5}$ Pa. The work involved in the process is $36.50$ J. Calculate the final volume.

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Solution

According to first law of thermodynamics

$W o r k (W) = n R T l n \frac{V_{2}}{V_{1}}$

$W = - 2.303 n R T l o g \frac{V_{2}}{V_{1}}$

$36.50 = - 2.303 \times 1 \times 8.314 \times l o g \frac{V_{2}}{500}$
after solving equation we get

$V_{2} = 6.2041 c m^{3}$

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Similar questions

Q. One mole of an ideal gas is compressed from

500

c m^{3}

against a constant pressure of

1.216 \times 10^{3}

Pa. The work involved in the process is

35.50 J

. Calculate the final volume.

Q. Calculate the work done when

5

moles of an ideal gas at

300

K are expanded isothermally from an initial pressure of

4 \times 10^{5}

Pa to a final pressure of

1 \times 10^{5}

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10^{5}

(5 \times 8.314 \times 300 = 12470)

Q. An ideal gas at pressure 2.5 × 10⁵ Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure (b) the final temperature and (c) the work done by the gas in the process. Take γ = 1.5

Q. Three mole of an ideal gas

(C_{p} = 7 / 2 R)

P_{A}

and temperature

T_{A}

is isothermally expanded to twice its initial volume. It is then compressed at constant pressure to its original volume. Finally the gas is compressed at constant volume to its original pressure

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curves for complete process.
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Q. Calculate the work done by adiabatic compression of one mole of an ideal gas (monoatomic) from an initial pressure of 1 atm to final pressure of 2 atm. Initial temperature =300 K.
(a) If the process is carried out reversibly
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Compute the final volume reached by gas in two cases and describe the work graphically .

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