Answer:
184.62 ml
Explanation:
Let
and
be the initial and
and
be the final pressure, volume, and temperature of the gas respectively.
Given that the pressure remains constant, so
...(i)
= 200 ml
K
K
From the ideal gas equation, pv=mRT
Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.
For the initial condition,

For the final condition,

Equating equation (i), and (ii)

[from equation (i)]

Putting all the given values, we have

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.
Answer:
so the answer that you get isn't wrong? i dont know
Answer: 1.48 atmosphere
Explanation:
Pressure in kilopascal = 150
Pressure in atmosphere = ?
Recall that 1 atmosphere = 101.325 kilopascal
Hence, 1 atm = 101.325 kPa
Z atm = 150 kPa
To get the value of Z, cross multiply
150 kPa x 1 atm = 101.325 kPa x Z
150 kPa•atm = 101.325 kPa•Z
Divide both sides by 101.325 kPa
150 kPa•atm/101.325 kPa = 101.325 kPa•Z/101.325 kPa
1.48 atm = Z
Thus, 150 kPa is equivalent to 1.48 atmospheres
Answer:
if the force applied increases
if the area of contact increases
Explanation: