A sealed container holding 0.0255 L of an ideal gas at 0.991 atm and 69 °C is placed into a refrigerator and cooled to 41 °C with no change in volume. Calculate thefinal pressure of the gas.
General guidance Concepts and reasonAs per the Gay-Lussac\u2019s Law, the relation between pressure and temperaute can be represented asPT=C\\frac = CTP\u200b=C Where, P is pressure, T is temperature and C is a constant. In the given question, change is pressure is to be calculated with change in temperature using Gay-Lussac\u2019s Law. When temperature and pressure are changed from one condition to another condition,the equation becomesP1T1=P2T2P2=P1T2T1\\begin\\\\\\frac{{}}{{}} = \\frac{{}}{{}}\\\\\\\\ = \\frac{{}}{{}}\\\\\\endT1\u200bP1\u200b\u200b=T2\u200bP2\u200b\u200bP2\u200b=T1\u200bP1\u200bT2\u200b\u200b\u200b FundamentalsGay-Lussac\u2019s law states that at constant volume (V) and for a given moles of gas (n), the ratio of pressure (P) to temperature (T) remains constant. Step-by-step Step 1 of 2 Convert temperature T1T1\u200b and T2T2\u200b into kelvin from degree Celsius as follows:T1=69\u2218C+273.15=342. 15K\\begin\\\\ = 69^\\circ {\\rm} + {\\rm}{\\rm{. 15}}\\\\\\\\ = {\\rm}{\\rm{. 15 K}}\\\\\\endT1\u200b=69\u2218C+273.15=342.15K\u200bAnd,T2=41\u2218C+273.15=314. 15K\\begin\\\\ = 41^\\circ {\\rm} + {\\rm}{\\rm{. 15}}\\\\\\\\ = 314. 15{\\rm{ K}}\\\\\\endT2\u200b=41\u2218C+273.15=314. 15K\u200b The temperature value should be in kelvin; therefore, conversion of unit is required. You need to convert the temperature into the kelvin unit. [Hint for the next step]Substitute the given values of temperature and pressure in the Gay Lussac\u2019s law formula for the final pressure. Step 2 of 2 Substitute P1P1\u200b as 0. 991 atm, T1T1\u200b as 342. 15 K and T2T2\u200b as 314. 15 K in the above equation as shown below:P2=P1T2T1=(0.991atm)(314.15K)342.15K=0. 909atm\\begin\\\\ = \\frac{{}}{{}}\\\\\\\\ = \\frac{{\\left( }} \\right)\\left( {{\\rm}{\\rm{. 15 K}}} \\right)}}{}}}\\\\\\\\ = 0. 909{\\rm{ atm}}\\\\\\endP2\u200b=T1\u200bP1\u200bT2\u200b\u200b=342.15K(0.991atm)(314.15K)\u200b=0. 909atm\u200b The final pressure is 0. 909 atm. The temperature is directly proportional to the pressure. The pressure goes down when the temperature goes down. As shown above, this decrease is calculated. Make sure that the correct units are used in the calculation. Answer The final pressure is 0. 909 atm. Answer only The final pressure is 0. 909 atm.
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