At stp,a certain mass of gas occupied a volume of 790cm3, find the temperature at which the gas occupied 1000cm3 and has a pressure of 728mmhg
To solve this problem, we can use the combined gas law equation, which relates the initial and final states of a gas in terms of pressure, volume, and temperature:
P1 × V1 / T1 = P2 × V2 / T2
We are given the initial volume (V1) and the final volume (V2), and we need to determine the final temperature (T2). Since the problem does not give us any information about the initial pressure (P1), we can assume that it is constant, and thus the P1 term cancels out on both sides of the equation:
V1 / T1 = V2 / T2
The problem states that the gas initially occupies a volume of 790 cm³ at standard temperature and pressure (STP). STP conditions correspond to a temperature (T1) of 273.15 K (0°C) and a pressure (P1) of 1 atm.
We are then given the final volume (V2) of 1000 cm³ and the final pressure (P2) of 728 mmHg. Our first step will be to convert this pressure to the same units as STP (1 atm), which is:
728 mmHg × (1 atm / 760 mmHg) = 0.9579 atm
Since we have now assumed that the pressure is constant, we can ignore this value and solve for the final temperature (T2) using the combined gas law equation:
(V1 / T1) = (V2 / T2)
Plug in the values and solve for T2:
(790 cm³ / 273.15 K) = (1000 cm³ / T2)
Cross multiply and solve for T2:
T2 = (1000 cm³ × 273.15 K) / 790 cm³
T2 = 346140 K / 790
T2 = 438 K
Therefore, the final temperature at which the gas occupied a volume of 1000 cm³ and had a pressure of 728 mmHg is approximately 438 K.