ThermodynamicsMedium
Question
2 moles of an ideal monoatomic gas undergoes reversible expansion from (4 L, 400 K) to 8 L such that TV2 = constant. The change in enthalpy of the gas is
Options
A.−1500R
B.−3000R
C.+1500R
D.+3000R
Solution
$T_{1}V_{1}^{2} = T_{2}V_{2}^{2} \Rightarrow T_{2} = T_{1}\left( \frac{V_{1}}{V_{2}} \right)^{2} = 400 \times \left( \frac{4}{8} \right)^{2} = 100\text{ K}$
Now, $\Delta H = n.C_{p,m}.\left( T_{2} - T_{1} \right) = 2 \times \frac{5}{2}R \times (100 - 400) = - 1500\text{ R}$
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