Question

Find the volume of a gas at STP, if its volume is 80.0 mL at 109 kPa and -12.5°C.​

Answer 1

Answer:

= 913.84 mL

Explanation:

Using the combined gas laws

P1V1/T1 = P2V2/T2

At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.

V1 = 80.0 mL

P1 = 109 kPa

T1 = -12.5 + 273 = 260.5 K

P2 = 10 kPa

V2 = ?

T2 = 273 K

Therefore;

V2 = P1V1T2/P2T1

     = (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)

     = 913.84 mL

Answer 2

Answer: The volume when the pressure and temperature has changed is 90.21 mL

Explanation:

At STP:

The temperature at this condition is taken as 273 K

The pressure at this condition is taken as 1 atm or 101.3 kPa.

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=101.3kPa\\V_1=?mL\\T_1=273K\\P_2=109kPa\\V_2=80.0mL\\T_2=-12.5^oC=[-12.5+273]K=260.5K$

Putting values in above equation, we get:

$\frac{101.3kPa\times V_1}{273K}=\frac{109kPa\times 80}{260.5K}\\\\V_1=\frac{109\times 80\times 273}{101.3\times 260.5}=90.21mL$

Hence, the volume when the pressure and temperature has changed is 90.21 mL