Question

The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K (R = 2 cal/mol-degree)

• A. 360 cal
• B. 720 cal
• C. 800 cal
• D. 1000 cal

Answer 1

Answer: (B)

720 cal

Answer 2

q = 0, $\Delta$U = W

nCV(T2–T1) = – Pext(V2–V1)

n$\frac{3}{2}$R (T2 300) = – 2 $\left( \frac{nRT_{2}}{2} - \frac{nR \times 300}{1} \right)$

$\Rightarrow$ $\frac{3}{2}$ (T2–300) = (600–T2) $\Rightarrow$ T2 = 420 K

W = nCV(T2–T1) = 2 ´ $\frac{3}{2}$ ´ 2 (420 – 300) = 720 cal.