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Vitek1552 [10]
3 years ago
10

A volume of 0.550 L of gas is contained at 23.0°C. What is the new volume of the gas if it is heated to 57.°C. Assume that the p

ressure of the gas remains constant. Work must be shown to earn credit.
Chemistry
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

0.61 L

Explanation:

T1 = 23.0 +273 = 296K

T2 = 57+273 = 330 K

V2 = (V1×T2)/T1

V2 = (0.55 × 330)/296

V2 = 0.61 L

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Using the equation for reversible work:

W = -P*(V_{2} - V_{1})

Where:

W is the work done (J) = -287 J.

Since the gas did work, therefore W is negative.

P is the pressure in atm = 1.90 atm.

However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K);  R = 0.0821 (L*atm)/(mol*K)

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(-287 J)*0.00987 (L*atm)/J = -1.9 atm*(V_{2} - 0.350) L

V_{2} = [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L

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