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2 years ago

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A gas at 110atm and 303K filled a container of 2L. If the temperature is raised to 353 K and the pressure is increased to 440atm

, what is the new volume
Question 4 options:

5.8 L

0.58 L

58 L

10.6 L

1 answer:

Sveta_85 [38]2 years ago

4 0

Answer:

0.58 L

Explanation:

For this problem we need to simply use the ideal gas equation to create a proportional comparison for the initial information to the final information.

(P_1 * V_1) / T_1 = (P_2 * V_2) / T_2

Using this, we can solve for V_2 to find the new volume of the gas once pressure and temperature changes.

(P_1 * V_1) * T_2 / T_1 = (P_2 * V_2)

(P_1 * V_1) * T_2 / (T_1 * P_2) = V_2

Consider our givens:

P_1 = 110atm

T_1 = 303K

V_1 = 2L

P_2 = 440atm

T_2 = 353K

Now we simply plug in these values to the equation to find the new volume, V_2.

(P_1 * V_1) * T_2 / (T_1 * P_2) = V_2

(110atm * 2L) * 353K / (303K * 440atm) = V_2

77660 atm*L*K / 133320 K*atm = V_2

0.583 L = V_2

Hence, the new volume is 0.583 L.

Cheers.

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substituting (1) into (2)

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$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

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