A sample of argon has a volume of 63 dm3 and the pressure is 121 atm. If the initial temperature is 61.° C, the final volume is 32.7 L, and the final pressure is 83 atm, what is the final temperature of the argon in Kelvin? Round your answer to the nearest Kelvin.

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To find the final temperature of the argon in Kelvin, we can use the combined gas law. The combined gas law relates the initial and final conditions of a gas sample:

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

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature (in Kelvin)

Given the following values:
P1 = 121 atm
V1 = 63 dm^3
T1 = 61.° C
P2 = 83 atm
V2 = 32.7 L
T2 = ?

First, we need to convert the initial volume and final volume to liters, as the units must be consistent.
63 dm^3 = 63 L (since 1 dm^3 = 1 L)
32.7 L = 32.7 L

Now, let's convert the initial temperature from Celsius to Kelvin.
To convert from Celsius to Kelvin, we use the formula:
T(K) = T(°C) + 273.15

T1 = 61.° C + 273.15 = 334.15 K

Now, we can rearrange the combined gas law to solve for T2:
T2 = (P2 * V2 * T1) / (P1 * V1)

Plugging in the values:
T2 = (83 atm * 32.7 L * 334.15 K) / (121 atm * 63 L)

T2 ≈ 193.87 K

Rounding to the nearest Kelvin,
T2 ≈ 194 K

Therefore, the final temperature of the argon is approximately 194 Kelvin.