A sample of gas at 38.0 degrees C occupies a volume of 2.97 L and exerts a pressure of 3.14 atm. The ga...

A sample of gas at 38.0 degrees C occupies a volume of 2.97 L and exerts a pressure of 3.14 atm. The gas is heated to 118.0 degrees C and the volume is decreased to 1.04 L. Determine the new pressure exerted by the gas. (convert temperature to K before substituting in equation)

Answer:
We can use the equation
p1V1 / T1 = p2V2 / T2

T1 = 38 + 273 = 311 K
T2 = 118 + 273 = 391 K

3.14 x 2.97 / 311 = p2 x 1.04 / 391
0.00340 = p2 x 0.00266
p2 = 11.3 atm
you have pV =nRT

in condition(1) p1 V1= nRT1
in condition(2) p2V2= nRT2
so p2V2/p1V1= T2/T1 p2V2T1 =p1V1T2 ; p2=p1V1T2/V2T1

here T1 =38+273=311K T2 = 118+273=391K
V1 =2.97L V2 = 2.97-1.04= 1.93L

so p2 = 3.14*391*3.14/(1.93*311)=6.43 atm
the eqn u need is P1V1/T1 =P2V2/T2
PLUG IN THE NUMBERS YOURSELF LAZY A**!

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