Question

A) A helium-filled weather balloon has a volume of 593 L at 16.9°C and 752 mmHg. It is released and rises to an altitude of 7.34 km, where the pressure is 354 mmHg and the temperature is –27.1°C. The volume of the balloon at this altitude is L.

B) A sample of helium gas occupies a volume of 7.24 L at 52.0°C and 1.14 atm.

If it is desired to increase the volume of the gas sample to 9.74 L, while increasing its pressure to 1.43 atm, the temperature of the gas sample at the new volume and pressure must be °C.

C) A sample of helium gas occupies a volume of 6.15 L at 58.0°C and 319 torr.

If the volume of the gas sample is decreased to 4.56 L, while its temperature is increased to 112.0°C, the resulting gas pressure will be  torr.

Answer 1

Here, applying the Gas equation,

PV=nRT

P is the pressure of the gas and 'T' is the temperature (Kelvin) of the gas.

'V' is the volume of the gas and 'R' is the gas constant.

'n' is the number of moles of the gas and remains constant on both the conditions here (as not mentioned).

All the problems are given of the same type except we have to find different parameters in all three.

applying gas equation for initial state and final state separately i.e.

P₁V₁ = RT₁ for initial state and P₂V₂ = RT₂ for final state.

As P₁V₁/T₁= R   and   P₂V₂/T₂ = R.

In both the situations, the whole term on the left-hand side is equal to a constant. Therefore the problem can be solved out by equating the two terms.

A) Initially,  

V₁ = 593 L V₂ = ?

T₁ = 273+16.9 = 289.9 Kelvin   T₂ = 273-27.1 = 245.9 Kelvin

P₁ = 752 mmHg   P₂ = 354 mmHg

Then putting the values and equating them,

P₁V₁/T₁ =  P₂V₂/T₂

=

Calculating V₂ from here;

V₂ = 1068.51 Litre.

Thus the volume of the balloon at the altitude is 1068.51 litre.

B) Here ;

V₁ = 7.24 L V₂ = 9.74 L

T₁ = 273+52 = 325 Kelvin T₂ = ?

P₁ = 1.14 atm P₂ = 1.43 atm

Then putting the values and equating them,

P₁V₁/T₁ =  P₂V₂/T₂

calculating the value for T₂

0.0254 = 13.93/T₂

T₂ = 548.43 Kelvin = 275.43^oC

Thus the temperature of the gas is 275.43^oC.

C) Similarly proceeding,

V₁ = 6.15 L V₂ = 4.56 L

T₁ = 273+58 = 331 Kelvin T₂ = 273+112 = 385 Kelvin

P₁ = 319 torr   P₂ = ?

Then putting the values and equating them,

P₁V₁/T₁ =  P₂V₂/T₂

calculating the value for P₂ :

5.927 = (P₂ * 4.56 L)/ 385 K

P₂= 500.42 torr

Hence the pressure of the gas at final conditions is 500.42 torr.