please help need to get this work in for tomorrow and im lost
A sample of argon has a volume of 6.0 dm3 and the pressure is 0.91 atm. If the final
temperature is 32° C, the final volume is 5.7 L, and the final pressure is 800 mmHg, what
was the initial temperature of the argon?
general gas equation
P1V1/T1=P2V2/T2
change your temps to kelvins, solve for t1, notice dm^3 is same as Liter L, convert .91 atm to mmgh (P2=.91*760 mmHg
T1=P1V1T2/P2V2
Is that a solution?
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of temperature, pressure, and volume.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 is the initial pressure,
V1 is the initial volume,
T1 is the initial temperature,
P2 is the final pressure,
V2 is the final volume, and
T2 is the final temperature.
Now let's plug in the values we know into the equation:
P1 = 0.91 atm
V1 = 6.0 dm3
T2 = 32°C + 273.15 (converting Celsius to Kelvin) = 305.15 K
P2 = 800 mmHg * (1 atm/ 760 mmHg) = 1.05 atm
V2 = 5.7 L
Plugging these numbers into the equation:
(0.91 atm * 6.0 dm3) / (T1) = (1.05 atm * 5.7 L) / (305.15 K)
Now, let's solve for T1:
T1 = (0.91 atm * 6.0 dm3 * 305.15 K) / (1.05 atm * 5.7 L)
Calculating this value will give you the initial temperature, which you can convert back to °C if necessary.
Remember to double-check your units and make sure they are consistent throughout the calculations.