Question

Calculate the temperature at which the volume of a given mass of gas gets reduced to 3/5th
of original volume at 10°C without any change in pressure.
SOLUTION​

Answer 1

Answer:

Given that original volume V

1

= V and final volume V

2

=

5

3

V

Also T

1

= 273 + 10 =283 K

Applying Charle's law;

T

1

V

1

=

T

2

V

2

T

2

=

5

3×283

= 169.8 K

Thus T

2

= 169.8 - 273 = -103.2

∘

C