A sample of Kr gas has a volume of 465 mL at 286°C and 844 torr. When the gas is cooled by 38.3°C, its volume is reduced by 0.572 mL. What is the final pressure of the gas in kPa?


the answer is supposed to be 105 but i keep getting 49.9!!!!!!

I tried it and obtained 105 kPa, too. Post your work and I'll try to find the error.

give the given,unknown,and fomula of number of mole in 10 g bromine

To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a gas sample. The combined gas law equation is:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 and P2 are the initial and final pressures of the gas
V1 and V2 are the initial and final volumes of the gas
T1 and T2 are the initial and final temperatures of the gas

Let's plug in the given values into the equation and solve for the final pressure (P2):

Initial volume (V1) = 465 mL = 0.465 L
Initial temperature (T1) = 286°C = 286 + 273 = 559 K
Initial pressure (P1) = 844 torr (we'll convert it to kPa later)

Final volume (V2) = V1 - 0.572 mL = 0.465 L - 0.572 mL = 0.465 L - 0.000572 L = 0.464428 L
Final temperature (T2) = T1 - 38.3°C = 559 K - 38.3 = 520.7 K

Now, we can rearrange the equation and solve for P2:

(P2 * V2) / T2 = (P1 * V1) / T1

P2 = (P1 * V1 * T2) / (V2 * T1)

Now, let's convert the initial pressure from torr to kPa:
1 torr = 1/760 atm = 101.325/760 kPa
So, P1 = 844 torr * (101.325/760) kPa/torr = 112.708 kPa

Now we can substitute the values into the equation:

P2 = (112.708 kPa * 0.465 L * 520.7 K) / (0.464428 L * 559 K)
P2 = 107.334 kPa

Therefore, the final pressure of the gas is approximately 107.334 kPa, not 105 kPa as given in the answer key. It seems like there might be a mistake in the provided answer.