# 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?

3.68 liters to three significant numbers.

This issue pertains to Charles Law.

thus

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You can use the combined gas law to solve this problem. The combined gas law equation is:

( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} )

Given: (V_1 = 3.50 , \text{L}) (T_1 = 25.0 + 273.15 , \text{K}) (T_2 = 40.0 + 273.15 , \text{K}) (P_1 = P_2 = 1.00 , \text{atm})

Substituting the values into the equation and solving for (V_2):

( \frac{(1.00 , \text{atm})(3.50 , \text{L})}{(25.0 + 273.15) , \text{K}} = \frac{(1.00 , \text{atm})(V_2)}{(40.0 + 273.15) , \text{K}} )

( V_2 = \frac{(1.00 , \text{atm})(3.50 , \text{L})(313.15 , \text{K})}{(298.15 , \text{K})} )

( V_2 ≈ 3.77 , \text{L} )

Therefore, the new volume of the balloon in the hot room is approximately 3.77 L.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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