The change in the magnitude of the volume of an ideal gas when a small additional pressure ΔP is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity ΔT at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively.
If |ΔT| = C|ΔP| then value of C in (K/atm.) is _________.

Solution

Physics sir · 3 steps

Find Δ V both ways from PV = nRT, set them equal, and V cancels out.

1. Constant temperature. With T fixed, PV is constant, so P Δ V + V Δ P = 0. The size of the volume change is |Δ V| = V/P|Δ P|.
2. Constant pressure. With P fixed, V/T is constant, so (Δ V)/V = (Δ T)/T. The size of the volume change is |Δ V| = V/T|Δ T|.

   V/P|Δ P| = V/T|Δ T| ⇒ |Δ T| = T/P|Δ P|

3. So. C = T/P = (300 K)/(2 atm) = 150 K/atm, so C is 150.

The idea to keep. For a small change in an ideal gas, use fractional changes: (Δ V)/V = -(Δ P)/P at constant T, and (Δ V)/V = (Δ T)/T at constant P.

Answer from the NTA final answer key. Solution by PrepWiser.