Question

A sample of gas at P = 1011 Pa, V = 1.49 L, and T = 315 K is confined in a cylinder.

a) Find the new pressure if the volume is reduced to half of the original volume at the same temperature.
b) If the temperature is raised to 415 K in the process of part (a), what is the new pressure?
c) If the gas is then heated to 630 K from the initial value and the pressure of the gas becomes 3033 Pa, what is the new volume?

Answer 1

Ideal gas law pV=nRT

where p is pressure in Pa, V is volume in cubic metres, n is number of mol of gas, R is the gas constant and T is temperature in kelvins.

a) if volume is reduced to half, and at the same temperature, pressure would double since volume is inversely proportional to pressure. Thus ans is 2022 Pa

b) the product of pressure and volume is proportional to temperature.

To find the new value of pV,
(1011 x 1.49) x ( 415/315)=1984.6 PaL (no need to convert L to cubic metres since we are just working in proportions)

Since volume reduced to half, the new value of pressure is,
1984.6/(1.49/2)=1478.5 Pa

c) Here the temperature doubles and the pressure triples, working with proportionality, we know that \[V \alpha T/p\]

so the overall result on volume is that it will volume will change by a factor of 2/3.
Thus new volume is (1.49)(2/3)= 0.99 L