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A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?

Use StartFraction V subscript 1 over T subscript 1 EndFraction equals StartFraction V subscript 2 over T subscript 2 EndFraction..

2.19 L
3.33 L
3.68 L
5.60 L

User Cgogolin
by
3.7k points

2 Answers

4 votes

Answer:

3.676 L.

Step-by-step explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

Knowing that:

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

User Fpietka
by
3.8k points
7 votes

Answer:

C- 3.68

Step-by-step explanation:

EDGE 2021

User Yamm
by
4.3k points