Question

Two moles of an ideal monoatomic gas are allowed to expand adiabatically and reversibly from 300 K and 200 K. The work done in the system is ($C_v=12.5J/K/mol$)

• A. $-12.5kJ$
• B. $-2.5kJ$
• C. $-6.25kJ$
• D. $500kJ$

Answer 1

The correct option is B $-2.5kJ$

$C_v=12.5J/K/mol$

$\gamma =C_p/C_v=(8.314+12.5)/12.5=1.665$

or an adiabatic process of ideal gas equation we have
PV^γ= K (Constant) (14)
γ = C_p/C_v

Suppose in an adiabatic process pressure and volume of a sample of gas changs from (P₁, V₁) to (P₂, V₂) then we have

P₁(V₁)^γ=P₂(V₂)^γ=K

Thus, P = K/V^γ

Work done by gas in this process is
W = -∫PdV

where limits of integration goes from V₁to V₂

Putting for P=K/V^γ, and integrating we get,

W = -(P₁V₁-P₂V₂)/(γ-1)

W = -nR(T1-T2)/(γ-1)

= -2*8.314(300-200)/(1.665-1)

=- 2500J=2.5kJ

Hence, option B is correct