Three moles of a gas (closed system) that obeys the equation of state P = (nRT/V

Question

Three moles of a gas (closed system) that obeys the equation of state P = (nRT/V) – (n2a/V2) undergo a reversible, isothermal expansion from an initial volume of V1 to a final volume of 15 V1 and thereby do –20,500 J of work. The temperature is 310 K and a = 3.87 L2 bar mol-2 for this gas. Calculate the initial volume V1 and the final pressure P2 for this change of state.

8.48 L, 40.6 bar

0.0784 L, 40.6 bar

0.821 L, 6.05 bar

0.821 L, 0.695 bar

0.0784 L, 6.05 bar

7.39 L, 0.403 bar

12.7 L, 0.403 bar

12.7 L, 0.606 bar

8.48 L, 0.606 bar

7.39 L, 0.695 bar

a.

8.48 L, 40.6 bar

b.

0.0784 L, 40.6 bar

c.

0.821 L, 6.05 bar

d.

0.821 L, 0.695 bar

e.

0.0784 L, 6.05 bar

f.

7.39 L, 0.403 bar

g.

12.7 L, 0.403 bar

h.

12.7 L, 0.606 bar

i.

8.48 L, 0.606 bar

j.

7.39 L, 0.695 bar

Explanation / Answer

answer is 0.0784 L of volume, 6.05 bar of pressure.

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