Question

A sample of argon gas is cooled and its volume went from 380 mL to 250 20 points

ml. If its final temperature was -55°C, what was its original temperature?

Answer 1

Applying the Laws of Thermodynamics we get as,

Step-by-step explanation:

V1/T1 = V2/T2

According to the law of Gaseous states.

so we get as,

V1 = 380 ml, V2 = 250 ml, T2 = -55°C ==> 273-55 ==> 218 Kelvin.

so now substituting we get as,

380/ T1 =250/218

==> 380*(218) /250 = T1

==> T1 = 331.36 ===>T1 = 331 K approx

So the Original Temperature was 331 K,

Option 1.) is correct.

Answer 2

The original temperature of the argon gas was approximately
`-0.0002899 L^2 K^-1.`

To find the original temperature of the argon gas, we can use the combined gas law equation:

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

where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

Since the pressure is not given, we can assume it remains constant for this problem. Therefore, we can rewrite the equation as:

(V1 / T1) = (V2 / T2)

We are given the initial volume (V1) as 380 mL and the final volume (V2) as 250 mL. We are also given the final temperature (T2) as -55°C.

Let's substitute these values into the equation and solve for the initial temperature (T1):

(380 mL / T1) = (250 mL / -55°C)

To get rid of the milliliter units, we can convert them to liters by dividing both sides of the equation by 1000:

(0.38 L / T1) = (0.25 L / -55°C)

Now, let's solve for T1:

T1 = (0.25 L / -55°C) * (0.38 L)

T1 = -0.25 L * 0.38 L / 55°C

T1 = -0.095 L^2 / 55°C

Since we are asked to provide the temperature in Kelvin (K), we need to convert from Celsius to Kelvin. The conversion formula is:

K = °C + 273.15

Let's convert -55°C to Kelvin:

T1 = -0.095 L^2 / (55°C + 273.15)

T1 = -0.095 L^2 / 328.15 K

Now we can calculate T1:

T1 = -0.095 L^2 / 328.15 K

T1 ≈ -0.0002899 L^2 K^-1