Question

A certain mass of a gas at standard temperature and pressure (STP) is expanded to 3 times its
o rigin a l v o lu m e un d e r ad iab a tic co n d itio n s. C a lc u la te th e re su ltin g te m p e ra tu re a n d pressure. (Take y = 1.4).

Answer 1

For adiabatic processes $pV^{\gamma}=\text{const}$ . Here $\gamma=1.4$ , therefore for initial and final conditions

$p_1V_1^{\gamma} = p_2V_2^{\gamma}$ . We know that $V_2 = 3V_1,$ therefore

$p_2 = p_1 \left(\dfrac{V_1}{V_2} \right)^{\gamma} = p_1 \left(\dfrac13 \right)^{\gamma} .$ Standard temperature and pressure imply the temperature $T=273.15\,\mathrm{K}, p = 10^5\,\mathrm{Pa}.$ So $p_1 = 10^5\,\mathrm{Pa}$ and

$p_2 = 10^5\,\mathrm{Pa}\cdot \left(\dfrac13 \right)^{1.4} \approx 2.15 \cdot10^4\,\mathrm{Pa}.$

According to the ideal gas law,

$p_1V_1 = \nu RT_1, \; p_2V_2 = \nu RT_2,$ so

$\dfrac{p_2V_2}{p_1V_1} = \dfrac{T_2}{T_1}, \;\; T_2 = T_1 \cdot \dfrac{p_2V_2}{p_1V_1} = 273.15\cdot \dfrac{2.15\cdot10^4\,\mathrm{Pa}\cdot3}{10^5\,\mathrm{Pa}} = 176\,\mathrm{K}.$