Final answer:
The new temperature of the gas at 6.50 atm can be calculated using Gay-Lussac's Law, which shows a direct relationship between pressure and temperature in gases. By setting up the proportion P1/T1 = P2/T2, and substituting in the values (including initial temperature converted to Kelvin), we find the final temperature to be 542.62 K or 269.47 °C.
Step-by-step explanation:
The student's question is about the relation between pressure and temperature of a gas, assuming volume and amount of moles remain constant. This scenario is described by Gay-Lussac's Law, which relates pressure and temperature directly, both of which must be in absolute units for the calculation (i.e., the temperature in Kelvins and pressure in atmospheres).
To find the new temperature at 6.50 atm, we convert the initial temperature from Celsius to Kelvin by adding 273.15 to 60 °C, giving us 333.15 K. With the initial pressure of 2.00 atm and the final pressure of 6.50 atm, we can set up the proportion P1/T1 = P2/T2 where P1 is the initial pressure, T1 is the initial temperature in Kelvins, P2 is the final pressure, and T2 is the temperature we are trying to find.
Thus, we have:
2.00 atm / 333.15 K = 6.50 atm / T2
T2 = (6.50 atm * 333.15 K) / 2.00 atm
T2 = 1085.2375 K / 2.00 atm
T2 = 542.62 K
Finally, if we wish to obtain the temperature in Celsius, we can subtract 273.15 from the result, yielding a final temperature of 269.47 °C.