Question

at 300K in thermal contact with surroundings expand isothermally from 1.0 L to 3.0 L against a constant pressure of 1.5 atm. In this process, the change in entropy of surroundings ($\Delta S_{surr}$) in JK⁻¹ is -x ×10⁻³. Find value c

Answer 1

1013

Answer 2

Solution

1. For an isothermal process, the first law gives ΔU = 0 so that

   $$Q = W.$$

2. The work done against a constant external pressure is

   $$W = P_{\text{ext}}\,\Delta V = 1.5\,\text{atm} \times (3.0 - 1.0)\,\text{L} = 1.5 \times 2 = 3.0\,\text{atm·L}.$$

3. Converting atm·L to Joules (1 atm·L = 101.325 J):

   $$Q = 3.0 \times 101.325 \approx 303.975\,\text{J}.$$

4. The change in entropy of the surroundings is given by

   $$\Delta S_{\text{surr}} = -\frac{Q}{T} = -\frac{303.975}{300} \approx -1.0133\,\text{J/K}.$$

5. Expressed in the form $-x \times 10^{-3}$ J/K, we have

   $$-1.0133\,\text{J/K} = -1013.3 \times 10^{-3}\,\text{J/K}.$$

   Hence,

   $$x \approx 1013.$$

Explanation (minimal):

• Compute $W=P_{\text{ext}}(V_f-V_i)$ and convert to Joules.
• Since $ΔU=0$ in an isothermal process, $Q=W$.
• Then use $ΔS_{\text{surr}}=-\frac{Q}{T}$ and express in the given form.