Question

A sample of nitrogen gas has a pressure of 6.58 kPa at 539K. If the volume does not

change, what will the pressure be at -62°C.

Answer 1

Explanation:

The formula to be used is:

P1/T1=P2/T2

At constant volume,we have

P1=6.58

P2=x(it wasn't given)

T1= 539

T2= -62°c(convert to kelvin:Celsius value + 273= {-62+273}= 211

6.58/539=X/211

Cross multiply and we have

X= 211×6.58/539

X=2.58 kPa

Answer 2

2.58 kPa

Explanation:

Initial Pressure of the nitrogen gas = 6.85 kPa

Initial temperature of the nitrogen gas 539 K

Final Pressure of the nitrogen gas = ???

Final Pressure of the nitrogen gas = -62°C = (273+(-62))K

= 211 K

∴ The equation that can be used to approach this question relating to pressure and temperature at a fixed constant volume is Gay-Lussac's Law of Combine Volume. The expression for Gay-Lussac's Law is given as:

`(P_1)/(T_1) = (P_2)/(T_2)`

`(6.58kPa)/(539K) =(P_2)/(211K)`

`P_2 = ((6.58kPa*211K))/(539K)`

`P_2= 2.58 kPa`

Thus, the pressure will be 2.58 kPa at -62°C