Date: [ t_(1)=0^circ mathrm(C) ] [ P_(1)=10^59 mathrm(~kg) ] [ V= ( const ) ] [ Delta P=3 cdot 10^5 mathrm(~m) ] 4) P= ? (P_(2))/(P_(1))= ? 6) T_(2)= ?

To solve this problem, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature, provided the volume remains constant.<br /><br />Given:<br />\( t_{1} = 0^{\circ} \mathrm{C} \)<br />\( P_{1} = 10^{59} \mathrm{~kg} \)<br />\( V = \text{const} \)<br />\( \Delta P = 3 \cdot 10^{5} \mathrm{~m} \)<br /><br />We need to find:<br />\( P =? \)<br />\( \frac{P_{2}}{P_{1}} =? \)<br />\( T_{2} =? \)<br /><br />Step 1: Convert the temperature from Celsius to Kelvin.<br />\( T_{1} = t_{1} + 273.15 = 0 + 273.15 = 273.15 \mathrm{~K} \)<br /><br />Step 2: Use the ideal gas law to find the pressure.<br />Since the volume is constant, we can use the formula \( \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \).<br /><br />Step 3: Substitute the given values into the formula.<br />\( \frac{10^{59}}{273.15} = \frac{P_{2}}{T_{2}} \)<br /><br />Step 4: Solve for \( P_{2} \).<br />\( P_{2} = \frac{10^{59} \cdot T_{2}}{273.15} \)<br /><br />Step 5: Calculate the value of \( P_{2} \) using the given \( \Delta P \).<br />\( \Delta P = P_{2} - P_{1} = 3 \cdot 10^{5} \)<br />\( P_{2} = P_{1} + \Delta P = 10^{59} + 3 \cdot 10^{5} \)<br /><br />Step 6: Substitute the value of \( P_{2} \) into the equation from Step 4.<br />\( 10^{59} + 3 \cdot 10^{5} = \frac{10^{59} \cdot T_{2}}{273.15} \)<br /><br />Step 7: Solve for \( T_{2} \).<br />\( T_{2} = \frac{(10^{59} + 3 \cdot 10^{5}) \cdot 273.15}{10^{59}} \)<br /><br />Step 8: Calculate the value of \( T_{2} \).<br />\( T_{2} = 273.15 + \frac{3 \cdot 10^{5}}{10^{59}} \)<br /><br />Step 9: Calculate the value of \( \frac{P_{2}}{P_{1}} \).<br />\( \frac{P_{2}}{P_{1}} = \frac{10^{59} + 3 \cdot 10^{5}}{10^{59}} \)<br /><br />Step 10: Calculate the value of \( P \).<br />\( P = P_{2} = 10^{59} + 3 \cdot 10^{5} \)<br /><br />Answer:<br />\( P = 10^{59} + 3 \cdot 10^{5} \)<br />\( \frac{P_{2}}{P_{1}} = \frac{10^{59} + 3 \cdot 10^{5}}{10^{59}} \)<br />\( T_{2} = 273.15 + \frac{3 \cdot 10^{5}}{10^{59}} \)