At a constant pressure, a 10.0 L volume of gas is cooled from 546 K to 273 K. What will be the final volume of this gas, assuming no liquefying occurs?

5.0 L

Explanation:

We must recall that pressure is only held constant when referring to Charles law. We then put down the parameters given in the question.

Initial volume V1= 10.0 L

Initial temperature T1= 546 K

Final temperature T2= 273 K

Final volume V2= the unknown

From the statement of Charles law;

V1/T1 = V2/T2

V1T2= V2T1

V2= V1T2/T1

V2= 10.0 × 273/ 546

V2= 5.0 L

Therefore, the new (final) volume is 5.0 L

Since the temperature is halved, the volume will also be halved.

The final volume is 5.0 L

Explanation:

Step 1: Data given

The pressure is constant

The initial volume of the gas = 10.0 L

The initial temperature = 546 K

The temperature is cooled to 273 K

Step 2: Calculate the final volume

V1/T1 = V2/T2

⇒with V1 = the initial volume = 10.0L

⇒with T1 = the initial temperature = 546 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the final temperature = 273 K

10.0 L / 546 K = V2 / 273 K

V2 = (10.0 L * 273 K) / 546 K

V2 = 5.0 L

Since the temperature is halved, the volume will also be halved.

The final volume is 5.0 L

the answer is option a.

um i think it might be a car?