Thermodynamics - Isothermal and Adiabatic Processes

One mole of an ideal gas at $27^\circ C$ is compressed isothermally until its volume is reduced to half its original volume. Calculate the work done on the gas. (Take $R = 8.31 J/mol\cdot K$ and $\ln 2 = 0.693$)

Given: $n = 1$, $T = 27 + 273 = 300 K$, $V_2 = \frac{V_1}{2}$. The work done by the gas is W = nRT \ln\left(\frac{V_2}{V_1} ight). Substituting values: $W = 1 \times 8.31 \times 300 \times \ln\left(\frac{1}{2}\right) = 1 \times 8.31 \times 300 \times (-0.693) \approx -1728.48 J$. Since the question asks for work done 'on' the gas, we take the magnitude: $W_{on} \approx 1728.48 J$.

In an isothermal compression, work is done on the system, which is reflected by the negative sign in the work done 'by' the gas formula. The energy transferred as work is dissipated as heat to keep the temperature constant.

A gas for which $\gamma = 1.5$ is suddenly compressed to $\frac{1}{4}$th of its initial volume. If the initial temperature is $27^\circ C$, find the final temperature.

Given: $\gamma = 1.5$, $V_2 = \frac{V_1}{4}$, $T_1 = 300 K$. Using the adiabatic relation $T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$, we have $T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}$. Substituting the values: $T_2 = 300 \times (4)^{1.5-1} = 300 \times 4^{0.5} = 300 \times 2 = 600 K$. Final temperature in Celsius is $600 - 273 = 327^\circ C$.

In a sudden compression (adiabatic), the work done on the gas increases its internal energy, leading to a significant rise in temperature because no heat is allowed to escape.