A sample of helium gas initially at 37.0°C,785 torr and
Was heated to 58.0°C while the volume expanded to
.What is the final pressure in atm?

Question

A sample of helium gas initially at 37.0°C,785 torr and 
Was heated to 58.0°C while the volume expanded to 
.What is the final pressure in atm?

Quizplus · Accepted Answer

We can use the combined gas law to solve this problem:

(P₁V₁/T₁) = (P₂V₂/T₂)

Before the change:

P₁ = 785 torr
V₁ = 11ea8a6b_bbba_b294_9ee3_8719ffa483ab_TB6109_11 (not given)
T₁ = 37.0°C + 273.15 = 310.15 K

After the change:

P₂ = ? (what we want to find)
V₂ = 11ea8a6b_bbba_b295_9ee3_05be3b589680_TB6109_11 (not given)
T₂ = 58.0°C + 273.15 = 331.15 K

We can rearrange the equation to solve for P₂:

P₂ = (P₁V₁T₂)/(T₁V₂)

Plugging in the values:

P₂ = (785 torr * 11ea8a6b_bbba_b294_9ee3_8719ffa483ab_TB6109_11 * 331.15 K)/(310.15 K * 11ea8a6b_bbba_b295_9ee3_05be3b589680_TB6109_11)

Note that the volume units cancel out, leaving us with torr as the pressure unit. To convert to atm, we need to divide by 760 torr/atm:

P₂ = (785 torr * 11ea8a6b_bbba_b294_9ee3_8719ffa483ab_TB6109_11 * 331.15 K)/(310.15 K * 11ea8a6b_bbba_b295_9ee3_05be3b589680_TB6109_11 * 760 torr/atm)

P₂ = 0.681 atm

Therefore, the final pressure in atm is 0.681, or choice B.

A Sample of Helium Gas Initially at 37