Question

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

Answer 1

Answer:

2.58 kPa  

Step-by-step 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:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

$\frac{6.58kPa}{539K} =\frac{P_2}{211K}$

$P_2 = \frac{(6.58kPa*211K)}{539K}$

$P_2= 2.58 kPa$

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

Answer 2

Answer:

Step-by-step 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