Question

A fixed quantity of gas at 21 °c exhibits a pressure of 752 torr and occupies a volume of 5.12 l. (a) calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. (b) calculate the volume the gas will occupy if the temperature is increased to 175 °c while the pressure is held constant.

Answer 1

A. PV=constant
752 torr = 0.989 atm
(0.989 atm)(4.38 L) = (1.88 atm)V
V = 2.30 L

B. T/V = constant
(294 K)/(4.38 L) = (448 K)/V
V = 6.67 L


Hoped this helped ( I believe it's right, sorry if it's not )

Answer 2

Answer:

For a: The final volume of gas is 2.69 L

For b: The final volume of gas is 2.69 L

Explanation:

• For a:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

Conversion factor used:  1 atm = 760 torr

$P_1=752torr=0.989atm\\V_1=5.12L\\P_2=1.88atm\\V_2=?L$

Putting values in above equation, we get:

$0.989atm\times 5.12L=1.88atm\times V_2\\\\V_2=\frac{0.989\times 5.12}{1.88}=2.69L$

Hence, the final volume of gas is 2.69 L

• For b:

To calculate the final volume of the gas, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=5.12L\\T_1=21^oC=(21+273)K=294K\\V_2=?\\T_2=175^oC=(175+273)K=446K$

Putting values in above equation, we get:

$\frac{5.12L}{294K}=\frac{V_2}{446}\\\\V_2=\frac{5.12\times 446}{294}=7.77L$

Hence, the final volume of gas is 7.77 L