Helium (He), a monatomic gas, fills a 0.010 m3 container. The pressure of the gas is 6.2×105Pa. How long would a 0.25 hp engine have to run (1 hp = 746W) to produce an amount of energy equal to the internal energy ofthis gas?

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avissidep

Answered question

2021-01-16

Helium (He), a monatomic gas, fills a 0.010 m3 container. The pressure of the gas is

6.2 \times 105 P a

. How long would a 0.25 hp engine have to run (1 hp = 746W) to produce an amount of energy equal to the internal energy ofthis gas?

Answer & Explanation

Aniqa O'Neill

Skilled2021-01-17Added 100 answers

Helium (He), a monatomic gas, fills a (V = 0.010 m3)container.
The pressure of the gas ( $P = 6.2 \times 105 P a$ ) .
The power =0.25 hp = 186.5 W engine have to run (1hp = 746 W) to produce an amount of energy equal to the internalenergy of this gas
So, Thechange in internal energy ( ? U ) = $(\frac{3}{2}) \times n \times R \times T$
But $P \times V = n \times R \times T$
So, ? U = $(\frac{3}{2}) \times P \times V$
Again $P o w e r \times t i m e = U$
$P o w e r \times t i m e$ = $(\frac{3}{2}) \times P \times V$
Calculate the value of time

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