Question

A sample of a gas at 15°C and 2.50 atm pressure has a volume of 4.5 L. The pressure is lowered to 0.85 atm and the volume decreases to 2.5 L. What is the final temperature of the gas in K.

Answer 1

Given:

$P_{1} = \text{2.50 atm}$

$T_{1} = 15^{\circ}\text{C + 273 = 288 K}$

$V_{1} = \text{4.5 L}$

$P_{2} = \text{0.85 atm}$

$V_{2} = \text{2.5 L}$

Unknown:

$T_{2}$

Solution:

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

$T_{2} = T_{1} \times \dfrac{P_{2}}{P_{1}} \times \dfrac{V_{2}}{V_{1}}$

$T_{2} = \text{288 K} \times \dfrac{\text{0.85 atm}}{\text{2.50 atm}} \times \dfrac{\text{2.5 L}}{\text{4.5 L}}$

$\boxed{T_{2} = \text{54.4 K}}$

#ILoveChemistry

#ILoveYouShaina