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3 years ago

A gas has a pressure of 50.0 mmHG at a temperature of 540 K. What will be the temperature if the pressure goes down to 3 mmHg?

Chemistry

2 answers:

slamgirl [31]3 years ago

6 0

Answer:

32.4 K

Explanation:

Initial pressure P1= 50.0mmHg

Initial temperature T1= 540K

Final temperature T2=????

Final pressure P2= 3 mmHg

From

P1/T1= P2/T2

P1T2= P2T1

T2= P2T1/P1

T2= 3×540/50.0= 1620/50.0

T2= 32.4 K

Alborosie3 years ago

5 0

Answer:

$T_2=32.4K$

Explanation:

Hello,

In this case, we use the Gay-Lussac's law which allows us to understand the pressure-temperature relationship as a directly proportional relationship:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Hence, we solve for the final temperature as shown below:

$T_2=\frac{T_1P_2}{P_1}= \frac{540K*3mmHg}{50.0mmHg} \\\\T_2=32.4K$

Best regards.

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nirvana33 [79]

Answer:

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Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

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a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

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