Physics, 07.12.2019 01:31 hebrew1148

During the time 0.332 mol of an ideal gas undergoes an isothermal compression at 26.4 c, 383 j of work is done on it by the surroundings. the final pressure is 1.87 atm. what was the intial pressure?

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Answer from: jerry7792

$P_{i}$ = 2.97 atm

Explanation:

The work in thermodynamic processes has several expressions, in the case of isothermal processes the work is

W = n RT ln ( $V_{i}$ / $V_{f}$ )

Where R is the constant of the ideal gases with a value of 8.314 J / mol K and T the absolute temperature

Let's look for the volume ratio

T = 26.4 +273.15 = 299.55k

Ln ( $V_{i}$ / $V_{f}$ )

= W / n RT

Ln( $V_{i}$ / $V_{f}$ )

= 383 / (0.332 8.314 299.55)

Ln( $V_{i}$ / $V_{f}$ )

= 0.4632

( $V_{i}$ / $V_{f}$ ) = e 0.4632

( $V_{i}$ / $V_{f}$ ) = 1,589

Let's use the ideal gas equation for two volumes

$P_{i} V_{i}$ = n R T

$P_{i} V_{i}$ = n RT

The temperature is the same because it is an isothermal process, this equation is the same

$P_{i} V_{i}$ = $P_{f} V_{f}$

Vi / Vf = $P_{f}/P_{i}$

$P_{i}$ = $P_{f}$ $V_{i}$ / $V_{f}$

$P_{i}$ = 1.87 1,589

$P_{i}$ = 2.97 atm

Answer from: Quest

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Answer from: Quest

Would like to . could you translate to english