A mole of gas at 0 degrees C and 760 mmHg occupies 22.41 L. What is the at 20 degrees C and 760 mmHg?
T1= 273 K
T2= 293 K
V1= 22.41 L
V2= ?
22.41 L * 293 K/ 273 K =
ANSWER:
24.05 L
Is this the correct answer
yes
Yes, the calculation you performed is correct. To solve this problem, you used the combined gas law, which states that the initial volume (V1) divided by the initial temperature (T1) is equal to the final volume (V2) divided by the final temperature (T2), assuming the pressure is constant.
By rearranging the formula, you can solve for V2:
V2 = V1 * (T2 / T1)
Plugging in the given values:
V2 = 22.41 L * (293 K / 273 K)
After performing the multiplication and division, you correctly obtained the answer as 24.05 L. Therefore, the volume of the gas at 20 degrees C and 760 mmHg would be approximately 24.05 L.