Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 19.3°C to 60.0°C.

Question

Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 19.3°C to 60.0°C.

Answer 1

We’re using Gay-Lussac’s Law, which states that for a constant volume of gas, pressure and temperature are directly proportional:

P1/T1=P2/T2

P1=1.00 atm

T1=19.3∘C+273.15=292.45 K

P2=?

T2=60.0 C+273.15K=333.15K

To solve for P2 we need to rearrange the equation to isolate the unknown:

P2=(P1)(T2)/(T1)

P2=(1.00 atm)(333.15K)/(292.45K)

P2=1.139 atm

The question is asking us to find the change in pressure so we need to subtract the final pressure-initial pressure

P2-P1=1.139 atm-1.00 atm= 0.139 atm

Answer 2

We are given an initial pressure of 1.00 atm (P1) and an initial temperature of 19.3ºC (T1). We area asked to find the final pressure (P2) when the final temperature is 60.0ºC(T2).

Since the volume is constant, we can use Gay-Lussac's Law (Amonton's Law):

P1/T1 = P2/T2

P1 = 1.00 atm

T1 = 19.3ºC + 273.15 = 292.45K

P2 = ?

T2 = 60.0ºC +273.15 = 333.15K

Solving for P2 we have...

P2 = (P1)(T2) / T1

P2 = (1.00 atm)(333.15K) / 292.45K

P2 = 1.139 atm

Since the question asks for the change in pressure, we will subtract the original pressure from the final pressure to obtain the change in pressure

Change in pressure = 1.139 atm - 1.000 atm

Change in pressure = 0.139 atm