Question

A 25.0-L sample of gas is compressed by a constant pressure of 1.267*10^5 Pa. If 328 J of work was done on the gas. what will be the final volume of the gas? (1 L atm= 101.325 J)

Answer #1

Given data:

w = 328 J

101.325 J = 1 L atm

So, 328 J = 328/101.325 = 3.237 L atm

w = 3.237 L atm

P = 1.267 x 105 Pa = (1.267 x 10^{5}) / (1.01325 x
10^{5}) = 1.250 atm ………. ( 1.01325 x 10^{5} Pa = 1
atm)

V_{1} = 25.0 L

V_{2} = ?

Work done in compression is given as,

w = - P.ΔV ………………… (Gas is compressed)

w = P (V_{2}-V_{1})

Let us put all known values,

3.237 = -1.250 (V_{2} – 25)

(V_{2} – 25) = -3.237/1.250

(V_{2} – 25) = -2.6

V_{2} = –2.6 + 25

V_{2} = 22.4 L

Final volume of gas will be 22.4 L.

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