Explanation:
(a) The given data is as follows.
Pressure on top (
) = 140 bar =
(as 1 bar =
)
Temperature =
= (15 + 273) K = 288 K
Density of gas = 


= 0.4548

=
= 
Hence, pressure at the natural gas-oil interface is
.
(b) At the bottom of the tank,

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]
= 
= 309.8 bar
Hence, at the bottom of the well at
pressure is 309.8 bar.
Answer: 400K
Explanation:
Given that,
Original volume of balloon V1 = 3.0L
Original temperature of balloonT1 = 27°C
Convert the temperature in Celsius to Kelvin
(27°C + 273 = 300K)
New volume of balloon V2 = 4.0L
New temperature of balloon T2 = ?
Since volume and temperature are given while pressure is constant, apply the formula for Charle's law
V1/T1 = V2/T2
3.0L/300K = 4.0L/T2
To get the value of T2, cross multiply
3.0L x T2 = 4.0L x 300K
3.0LT2 = 1200LK
Divide both sides by 3.0L
3.0LT2/3.0L = 1200LK/3.0L
T2 = 400K
Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.