A sample of helium gas has a volume of 900 milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the temperature is changed to 336 K and the volume is decreased to 450 milliliters?

Question

Serenity Allen · Accepted Answer

#(P_1V_1)/T_1 = (P_2V_2)/T_2# This is the combined gas law. And it saves us from tedious mucking about with gas constants and having to standardize units.

So following the gas law: #P_2 = (P_1V_1)/T_1 xx T_2/V_2 = (2.50*atmxx900xx10^(-3)cancelLxx336cancelK )/(298cancelKxx450xx10^(-3)cancelL)# #=# #>5*atm ?# Since we are both compressing the gas and increasing its temperature, the pressure change makes sense.

Natalie Anton · Answer

undefined To solve for the new pressure, you can use the combined gas law formula:

$$ P_1V_1/T_1 = P_2V_2/T_2 $$

Where:
$ P_1 = 2.50 \, \text{atm} $ (initial pressure)
$ V_1 = 900 \, \text{mL} $ (initial volume)
$ T_1 = 298 \, \text{K} $ (initial temperature)
$ V_2 = 450 \, \text{mL} $ (final volume)
$ T_2 = 336 \, \text{K} $ (final temperature)

Plug in the values and solve for $ P_2 $:

$$ P_2 = \frac{P_1V_1T_2}{V_2T_1} $$

$$ P_2 = \frac{(2.50 \, \text{atm})(900 \, \text{mL})(336 \, \text{K})}{(450 \, \text{mL})(298 \, \text{K})} $$

$$ P_2 ≈ 4.47 \, \text{atm} $$

So, the new pressure when the temperature is changed to 336 K and the volume is decreased to 450 milliliters is approximately 4.47 atm.