Question

What is the correct numerical setup for calculating the final volume of the helium gas sample?

Answer 1

Answer:

This question is incomplete. But the completed question is below

A sample of helium gas is in a closed system with a movable piston. The volume of the  gas sample is changed when both the temperature and the pressure of the sample are increased.  Where the initial temperature of the gas is 200K, initial pressure is 2.0 atm, initial volume is 500 mL, the final temperature is 300K and the final pressure of the gas sample is 7.0 atm.

What is the correct numerical setup for calculating the final volume of the helium gas sample?

P₁V₁/T₁ = P₂V₂/T₂ will be used in calculating the final volume.

And the final volume is 214.29 mL

Explanation:

The general gas equation will be used to calculate the final volume. The general gas equation is

P₁V₁/T₁ = P₂V₂/T₂

From the question

P₁ = 2.0 atm

T₁ = 200K

V₁ = 500 mL

P₂ = 7.0 atm

T₂ = 300K

V₂ = unknown

Hence, we have

2 × 500/200 = 7 × V₂/300

V₂ = $\frac{2 * 500 * 300}{200 * 7}$

V₂ = 214.29 mL

Answer 2

The final volume of He gas sample has been 214.285 mL.

Helium has been assumed to be an ideal gas. The final volume of the gas with an increase in pressure has been given by:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

Where, the initial pressure of gas, $P_1=2\;\text{atm}$

The final pressure of gas, $P_2=7\;\text{atm}$

The initial volume of gas, $V_1=500\;\text{ml}$

The initial temperature of gas, $T_1=200\;\text K$

The final temperature of gas, $T_2=300\;\text K$

Substituting the values for the final temperature of He, $V_2$:

$\dfrac{2\;\times\;500}{200}=\dfrac{7\;\times\;V_2}{300}\\\\\dfrac{2\;\times\;500}{200}\;\times\;\dfrac{300}{7} = V_2\\\\V_2=214.285\;\rm ml$

The final volume of He gas has been 214.285 mL.

For more information about the volume of gas, refer to the link:

https://brainly.com/question/1323061