Question

A sample of 1.00 mol of N2 gas is expanded adiabatically from a volume of 10.00 dm3 and a temperature of 400 K to a volume of 20.00 - 3 -dm3. Assume that nitrogen is ideal, with Cv,m = 5R/2. (i) Find the final temperature if the expansion is carried out reversibly. (ii) Calculate the final temperature if the expansion is carried out with a constant external pressure of 1.00 atm. (iii) Find the final temperature if the gas expands into a vacuum. (iv) Find ∆U and w for each of the processes in the three parts, above. (v) For the reversible expansion in (i), show that the value of w obtained from the integral wrev = -[Integral]Pdv is the same as the value of w obtained from w = ∆U – q = ∆U.

Answer 1

(i) for reversible adibatic expansion :

$\frac{T_2}{T_1} = \left ( \frac{V_1}{V_2} \right )^{\gamma -1} \ \ \ \ \ \ \ \ \ \ \ \ ---eq. (1))$

where $\gamma = \frac{C_p}{C_v}$ and for diatomic molecule γ is 7/5.

putting the value in eq. (1)

$\frac{T_2}{T_1} = \frac{T_2}{400} = \left ( \frac{10}{20} \right )^{\frac{7}{5} -1} = \left ( \frac{1}{2} \right )^\frac{2}{5} \approx 0.76$

$\frac{T_2}{T_1} = \frac{T_2}{400} = \left ( \frac{10}{20} \right )^{\frac{7}{5} -1} = \left ( \frac{1}{2} \right )^\frac{2}{5} \approx 0.76$

$T_2 = 400 * 0.76 \approx 304 K$

= 304 K

(ii) first calculate the pressure for intial condition by using ideal gas eq. PV = nRT:

P*10 = 1*8.314*400