A container holds 50.0 mL of nitrogen at 25° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?

Question

A container holds 50.0 mL of nitrogen at 25° C and a pressure of 736 mm Hg.  What will be its volume if the temperature increases by 35° C?

Peyton Adams · Accepted Answer

For this problem we would use the equation for Charles' Law which compares the direct relationship between Volume and Temperature when Pressure is held constant.

Since there is no mention of a change in the pressure we must assume that pressure is held constant.

#V_i# = 50.0 mL
#T_i# = 25 C + 273 = 298 K
#V_f# = ???
#T_f# = 60 C + 273 = 333 K

Always change the temperature form Celsius to Kelvin for the gas laws.

Charles' Equation is

#V_i/T_i = V_f/T_f#

This can be rearranged algebraically to

#(V_iT_f)/T_i = V_f#

#((50.0 mL)(333K))/(298 K) = V_f#

#V_f# #= 55.9 mL #

I hope this was helpful.
SMARTERTEACHER

Ellie Andrews · Answer

undefined To solve this problem, you can use the combined gas law, which states: $P_1V_1/T_1 = P_2V_2/T_2$.

Given:
$V_1 = 50.0 \, \text{mL}$
$T_1 = 25°C + 273.15 = 298.15 \, \text{K}$
$P_1 = 736 \, \text{mm Hg}$
$T_2 = 25°C + 35°C + 273.15 = 333.15 \, \text{K}$

Rearranging the formula to solve for $V_2$, we get:

$V_2 = (P_1 \cdot V_1 \cdot T_2) / (P_2 \cdot T_1)$

Substitute the values and calculate:

$V_2 = (736 \, \text{mm Hg} \cdot 50.0 \, \text{mL} \cdot 333.15 \, \text{K}) / (736 \, \text{mm Hg} \cdot 298.15 \, \text{K})$

$V_2 \approx 58.6 \, \text{mL}$

So, the volume will be approximately 58.6 mL when the temperature increases by 35°C.