Question

For each of the following sets of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain the same.

(a) V = 2.06 L at 28°C; V = 3.41 L at WebAssign will check your answer for the correct number of significant figures. °C

(b) V = 105 mL at 221 K; V = WebAssign will check your answer for the correct number of significant figures. mL at 376 K

(c) V = 48.8 mL at 40.°C; V = WebAssign will check your answer for the correct number of significant figures. mL at 369 K

Answer 1

According to the Charle's law

For a gas , temperature and volume are directly proportional . . Here pressure and amount of gas is remain same .

$V \alpha T$

V = KT

Where V = volume of the gas

K = proportionality constant

T = temperature in kelvin

Or $\frac{V}{T}= K$

Applying Charle's Law in different temperature and volume for a gas

$\frac{V_{1}}{T_{1}}= K$ and   $\frac{V_{2}}{T_{2}}= K$

Hence

  $\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$

a) here V ₁ = 2.06 L

T_{​​​​​1} = 28°C = (28+273.15)K = 301.15K

V ₂ = 3.41 L

Using Charle's law

  $\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$

$\frac{2.06L}{301.15K }= \frac{3.41L}{T_{2}}$

T_{​​​​​​2} = 498.50K

Hence the temperature is (498.50-273.15)K = 225.35 °C

b) Here V_{​​​​​​1} = 105 mL

T_{​​​​​​1} = 221 K

V_{​​​​​​2} = unknown

T_{​​​​​​2} = 376 K

Using Charle's law

  $\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$

$\frac{105 mL }{221K}= \frac{V_{2}}{376K }$

$V_{2}=\frac{105 mL \times 376 K }{221K}$

V_{​​​​​2} = 178.64 mL

Hence the volume is 178.64 mL

C) here   

V_{​​​​​​1} = 48.8 mL

T_{​​​​​​1} = 40°C= (40+273.15) K = 313.15 K

V_{​​​​​​2} = unknown

T_{​​​​​​2} = 369 K

Using Charle's law

$\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$

  $\frac{48.8 mL }{313.15K}= \frac{V_{2}}{369K }$

  $V_{2}=\frac{48.8 mL \times 369 K }{313.15K}$

V_{​​​​​​2} = 57.50 mL

Hence the volume is 57.50 mL .