The work done by an ideal gas in an isothermal expansion from volume V_1 to volu
ID: 1623021 • Letter: T
Question
The work done by an ideal gas in an isothermal expansion from volume V_1 to volume V_2 is given by the formula: W = nRT ln(V_2/V_1) Standard atmospheric pressure (1 atm) is 101.3 kPa. If 1.0 L of He gas at room temperature (20 degree C) and 1.0 atm of pressure is compressed isothermally to a volume of 100 mL, how much work is done on the gas? A) 5.6 kJ B) 4.7 times 10^2 J C) 4.7 times 10^2 kJ D) 2.3 times 10^2 kJ E) 2.3 times 10^2 J A gas expands along path ABC. This work done by the gas in this expansion is A) 4.0 times 10^5 J B) 5.0 times 10^5 J C) 6.0 times 10^5 J D) 7.0 times 10^5 J E) 8.0 times 10^5 J An ideal gas initially at 50 degree C and pressure P_1 = 100 kPa occupies a volume V_1 = 3 L. It undergoes a quasi-static, isothermal expansion until its pressure is reduced to 50 kPa. How much work was done by the gas during this process? R = 8.314 J/mol K = 8.206 L atm/mol K. A) 116 J B) 208 J C) 256 J D) 304 J E) 416 JExplanation / Answer
11. W = n R T ln(V2 / V1)
W = Pi Vi ln(V2 / V1 )
= (101.3 x 10^3 x 10^-3 ) ln(0.10 / 1 )
= - 233.25 J
Work done on the gas = - work done by gas = 233.25 J
Ans(E)
12. work done = area under the curve
= 3 ( 0.2 x 1) + (0.2 x 1 / 2)
= 0.7 x 10^6 J = 7 x 10^5 J
Ans(D)
13. Work done = P1 V1 ln(P1 / P2)
= (100 x 10^3 x 3 x 10^-3) ln( 100 / 50)
= 208 J
ans(B)
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