# Answer to Question #172251 in General Chemistry for Kaylee Holbrook

Question #172251

A sample of hydrogen gas at 341 K experiences a change in volume from 5.49 L to 3.12 L. If its new pressure is 3.48 atm at 215 K, what was its original pressure in atmospheres?

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2021-03-17T06:39:31-0400

Q172251

A sample of hydrogen gas at 341 K experiences a change in volume from 5.49 L to 3.12 L. If its new pressure is 3.48 atm at 215 K, what was its original pressure in atmospheres?'

Solution:

Initial temperature of hydrogen gas is 341K.

Volume of sample of hydrogen gas changed from 5.49 L to 3.12L, so

V initial = 5.49 L and V final = 3.12 L.

New pressure is 3.48 atm at 215K, which means

P final = 3.48 atm and T final = 215 K.

We have to find the original pressure ( initial ) in 'atm',

P initial = unknown ; P final = 3.48 atm

V initial = 5.49 L ; V final = 3.12 L

T initial = 341 K ; T final = 215 K

We can use the combined gas equation here.

"\\frac{P_1\\space V_1}{T_1} = \\frac{P_2\\space V_2}{T_2}"

"\\frac{P_{initial}\\space V_{initial}}{T_{initial}} = \\frac{P_{final }\\space V_{final }}{T_{final }}"

Substitute all the given information in the formula, we have

"\\frac{P_{initial}\\space *\\space 5.49\\space L}{341\\space K} = \\frac{3.48\\space atm\\space *\\space 3.12\\space L }{215\\space K }"

"P_{initial}\\space * 0.01610 L\/K = 0.050500L.atm\/K"

divide both the side by 0.01610 L/K, we have

"\\frac{P_{initial}\\space *\\space 0.01610\\space L\/K}{0.01610\\space L\/K} = \\frac{0.050500\\space L.atm\/K}{0.01610\\space L\/K }"

"P_{initial} = \\frac{0.050500\\space L.atm\/K}{0.01610\\space L\/K } = 3.137 atm."

In the question all the quantities are given in 3 significant figure, so our final answer must also

be in 3 significant figure.

### Hence the Original (initial) pressure of sample of hydrogen gas will be 3.14 atm.

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