An ideal gas, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume.

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Solution

Q.
An ideal gas, ${\vec{C}}_{v} = \frac{5}{2} R,$ is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is __________ K (nearest integer). [ ${\vec{c}}_{v}$ is the molar heat capacity at constant volume] [2024]

Ans.
(274)

Work done (W) by gas against a constant pressure is given by:

$W = - P_{ext} (V_{f} - V_{i}) = - 1 (2 V - V) = - V$ ...(i)

Change in internal energy ( $∆ U$ ) for an ideal gas undergoing any process is:

$Δ U = n C_{v} (T_{f} - T_{i}) = n \times \frac{5}{2} R (T_{f} - 298)$ ...(ii)

By first law of thermodynamics:

$Δ U = q + W$

Since q = 0 for an adiabatic process,

$Δ U = W$

Put $∆ U$ from (ii) and W from (i)

$n \times \frac{5}{2} R (T_{f} - 298) = - V$ ...(iii)

Apply ideal gas equation at initial conditions,

$P_{i} V_{i} = n R T_{i}$

$5 V = n R \times 298$

$V = \frac{n R \times 298}{5}$ ...(iv)

Put V from (iv) in (iii)

$n \times \frac{5}{2} R (T_{f} - 298) = - \frac{n R \times 298}{5}$

$\frac{5}{2} (T_{f} - 298) = - \frac{298}{5}$

$T_{f} - 298 = - \frac{298 \times 2}{5 \times 5}$

$T_{f} = 274.16 \approx 274 K$

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