Question

Determine the pressure change when a constant volume of gas at 101.3 kPa is heated from 25oC to 30oC, providing the initial and final temperatures and pressures, along with the gas constant.

Answer 1

The pressure of a gas at constant volume increases from 101.3 kPa to 102.6 kPa when heated from 25°C to 30°C.

Step-by-step explanation:

To determine the pressure change when a constant volume of gas is heated, we can use the Gay-Lussac's law, which is a derivation of the ideal gas law when volume (V) and the amount of gas (n) are constant. The law relates pressure and temperature of a gas as follows: P1/T1 = P2/T2, where P is pressure, and T is temperature.

In this case, we have an initial pressure (P1) of 101.3 kPa, an initial temperature (T1) of 25°C, which is 298.15 K (since temperature in Kelvin is the Celsius temperature plus 273.15), and a final temperature (T2) of 30°C, which is 303.15 K. To find the final pressure (P2), we rearrange the equation to P2 = P1 × (T2/T1).

Substituting the values in, we get:
P2 =101.3 kPa × (303.15 K / 298.15 K)
P2 = 102.6 kPa

The final pressure after heating the gas to 30°C is 102.6 kPa; hence, the pressure increases by 1.3 kPa.