Question

a volume of hydrogen measure 1 cubic decimeter at 20 degree Celsius and at a pressure of half an atmosphere what will be its volume at 10 degree Celsius and at 700 mm pressure

Answer 1

Ideal gas equation,

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

V_{1} = 1 dm^{3} , P_{1} =0.5 atm, T_{1} = (20 + 273) K = 293 K\\

V_{2}= ?, P_{2} = 700 mmHg×$\frac{1 atm}{760mmHg}$ = 0.921 atm, $T_{2} = (10+ 273) K = 283 K$

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

=$\frac{0.5 atm(1 dm^{3})(283 K)}{(293 K)(0.921 atm)}$

= 0.524 $dm^{3}$

Answer 2

Answer : The volume at $10^oC$ and 700 mmHg pressure is, 0.5243 L

solution : Given,

Initial volume = $1dm^3=1L$

Initial temperature = $20^oC=273+20=293K$

Initial pressure = 0.5 atm

Final temperature = $10^oC=273+10=283K$

Initial pressure = $700mmHg=\frac{700}{760}=0.921atm$     $(1atm=760mmHg)$

Formula used :

Using ideal gas law,

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

where,

$P_1$ = initial pressure

$P_2$ = final pressure

$V_1$ = initial volume

$V_$ = final volume

$T_1$ = initial temperature

$T_2$ = final temperature

Now put all the given values in the above formula, we get

$\frac{(0.5atm)\times (1L)}{293K}=\frac{(0.921atm)\times V_2}{283K}$

By rearranging the terms, we get the final volume.

$V_2=0.5243L$

Therefore, the volume at $10^oC$ and 700 mmHg pressure is, 0.5243 L