Question

Calculate the fall in pressure of helium initially at 1600 Pa, when it is suddenly

expended to 8 times its original volume. Given γ= 5/3​

Answer 1

Answer:

Expanded suddenly → Adiabatic process

TV

Y−1

= constant

TV

5/3−1

= const

TV

2/3

= const

T

1

V

1

2/3

=T

2

V

2

2/3

⇒

T

2

T

1

=(

V

1

V

2

)

2/3

⇒

T

2

T

1

=(8)

2/3

⇒

T

2

(20+273)K

=4

⇒T

2

=

4

293

K

Answer 2

Answer:

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