Question

One kg of dry air at \(15~^\circ\mathrm{C}\) is isothermally compressed to one–tenth of its initial volume. The work done on the system is ______________________________________ kJ. (Round off to the nearest integer) [Use the gas constant for dry air \(R=287~\mathrm{J\,kg^{-1}\,K^{-1}}\).]

Hint:

Memorize: For isothermal ideal gas, \(W_{\text{on}}=mRT\ln\!\left(\dfrac{V_1}{V_2}\right)=mRT\ln\!\left(\dfrac{p_2}{p_1}\right)\). The result scales linearly with \(T\) and \(\ln\) of the compression ratio.

Answer 1

Correct Answer: 189

Step 1 (Process identification).
\(\Rightarrow\) Isothermal, ideal-gas compression \(\Rightarrow\) \(T=\) constant and \(pV=\text{const}\) for a fixed mass. The exact work for an isothermal ideal-gas process is obtained by integrating \(W_{\text{by}}=\int p\,\mathrm{d}V\) with \(p=\dfrac{mRT}{V}\):
\[ W_{\text{by}}=\int_{V_1}^{V_2}\frac{mRT}{V}\,\mathrm{d}V=mRT\ln\!\left(\frac{V_2}{V_1}\right). \]

Step 2 (Numerical values and sign convention).
Given \(m=1~\mathrm{kg}\), \(T=15^\circ\mathrm{C}=288.15~\mathrm{K}\), and \(V_2/V_1=0.1\). \[ W_{\text{by}}=(1)(287)(288.15)\ln(0.1). \] Compute the product: \(287\times 288.15=82{,}699~\mathrm{J\,kg^{-1}}\). \(\ln(0.1)=-2.302585\). Hence \[ W_{\text{by}}=82{,}699\times(-2.302585)=-190{,}400~\mathrm{J}. \]

Step 3 (Work on the system).
By the usual sign convention, compression gives negative work \emph{by} the system. The work \emph{on} the system is the negative of this value: \[ W_{\text{on}}=-W_{\text{by}}=190{,}400~\mathrm{J}=190.4~\mathrm{kJ}. \]

Step 4 (Unit and sanity check).
Isothermal compression by a factor of \(10\) typically gives \(W_{\text{on}}=mRT\ln(10)\approx (1)(287)(288)\times 2.303\approx 190~\mathrm{kJ}\) — consistent. Final Answer:
\[ \boxed{190~\text{kJ}} \]