With increasing pressure, the temperature dependence of internal pressure changes. It reflects the crossing point of the isotherms of the internal pressure. The internal pressure decreases with increasing temperature at pressures up to the crossing point and then it increases with the increase of temperature.

Examples of simple gas calculations

1. Calculate the final pressure.
2. p1 x V1 = p2 x V2
3. rearranging gives p2 = (p1 x V1) / V2
4. p2 = (101 300 x 5) / 2.8 = 180893 Pa.

Boyle’s Law – states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.

Charles’s law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant. It is a special case of the general gas law and can be derived from the kinetic theory of gases under the assumption of a perfect (ideal) gas.

The pressure is kept constant in Charles law. According to Charles law, the volume of the gas is proportional to the temperature if the pressure remains constant. Was this answer helpful?

According to Charles law, when the pressure remains constant, the volume of a fixed amount of gas is directly proportional to the temperature.

Answer. Answer: An ideal gas undergoes an adiabatic process obeying the relation PV^4/3 = constant.

So according to the question V1 is V and V2 is 5% less than the original volume at a constant temperature. We know that change in pressure is equal to the difference in the final pressure and the initial pressure. So, 5.26% of pressure must be increased in order to decrease the volume of the gas by 5%.

Answer. It is given that water is to be compressed by 0.10%. Therefore, the pressure on water should be 2.2 ×106 Nm–2.

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Thus , if the volume is increased by 10% , the pressure should decreases by 10%.

To decrease the pressure of the gas by 10% at constant temperature then what should the change in volume be? So V²/V¹= 1.111 x P¹. Therefore to decrease the pressure by 10% , you have to increase the volume by 1/9- times or by 11.11% .