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asambeis [7]
3 years ago
15

You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

w pressure be?
Chemistry
1 answer:
fenix001 [56]3 years ago
7 0

Answer:

P_2=1.90atm

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}

Whereas both n and R are cancelled out as they don't change, we obtain:

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

Thus, by solving for the final pressure, we obtain:

\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm

Best regards!

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