5. There is carbon dioxide at pressure p_(1)=6 bar and temperature t_(1)=527^circ C in a tank of vol- ume V=0.5m^3 . How will the temperature of the gas change if 436 kJ of heat are removed from it at constant volume?

To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.<br /><br />Given:<br />- Initial pressure, $p_{1}=6$ bar<br />- Initial temperature, $t_{1}=527^{\circ }C$<br />- Volume, $V=0.5m^{3}$<br />- Heat removed, $Q=-436$ kJ (negative because heat is removed)<br /><br />We need to find the final temperature, $t_{2}$.<br /><br />First, let's convert the initial temperature to Kelvin:<br />$t_{1} = 527^{\circ }C + 273.15 = 800.15 K$<br /><br />Next, we can use the ideal gas law to find the initial number of moles of the gas:<br />$p_{1}V = nRT_{1}$<br />$n = \frac{p_{1}V}{RT_{1}}$<br /><br />Now, we can use the first law of thermodynamics to find the change in internal energy:<br />$\Delta U = Q - W$<br />Since the volume is constant, $W = 0$, so $\Delta U = Q$<br /><br />The change in internal energy can also be expressed as:<br />$\Delta U = nC_{v}\Delta T$<br />where $C_{v}$ is the molar specific heat capacity at constant volume.<br /><br />Substituting the values, we get:<br />$nC_{v}\Delta T = -436 kJ$<br /><br />Now, we can solve for the change in temperature:<br />$\Delta T = \frac{-436 kJ}{nC_{v}}$<br /><br />Finally, we can find the final temperature:<br />$t_{2} = t_{1} + \Delta T$<br /><br />Note: The specific heat capacity, $C_{v}$, depends on the gas in question. For an ideal monatomic gas, $C_{v} = \frac{3}{2}R$, where $R$ is the universal gas constant. For an ideal diatomic gas, $C_{v} = \frac{5}{2}R$.