Question

A gas is allowed to expand in a insulated container against a constant external pressure of 10 atm from an initial volume 4.2 L to 5.2 L. The change in internal energy ΔE of the gas will be :

Answer 1

-1013 J

Answer 2

Solution:

Since the container is insulated (adiabatic process), there is no heat exchange ($q = 0$). Thus, from the first law of thermodynamics

$$\Delta E = q + w,$$

we have

$$\Delta E = w.$$

The work done by the gas when expanding against a constant external pressure is given by

$$w = -P_{\text{ext}}\,\Delta V,$$

where

$$\Delta V = V_f - V_i = 5.2\,\text{L} - 4.2\,\text{L} = 1.0\,\text{L}.$$

Converting pressure and volume to SI units:

• $P_{\text{ext}} = 10\ \text{atm} = 10 \times 101325\ \text{Pa}$.

• $1.0\,\text{L} = 1.0 \times 10^{-3}\,\text{m}^3$.

Thus,

$$w = -\left(10 \times 101325\,\text{Pa}\right) \times \left(1.0 \times 10^{-3}\,\text{m}^3\right) = -1013.25\,\text{J}.$$

Therefore,

$$\Delta E \approx -1013\,\text{J}.$$

Core Explanation:

An adiabatic expansion implies $q=0$, so $\Delta E = w$. Calculate work using $w = -P\Delta V$ with proper unit conversions yielding approximately $-1013$ J.