# Answer to Question #33615 in Molecular Physics | Thermodynamics for Nishant

Question #33615

1) What volume will a gas occupy at 740 mmpressure which at 1480 mm occupies 500cc?

2) At a given temperature the pressure of a gas reduces to 75% of its initial value and the volume increases by 40% of its initial value. Find temp. if the initial temp. was 10 degree celcius

2) At a given temperature the pressure of a gas reduces to 75% of its initial value and the volume increases by 40% of its initial value. Find temp. if the initial temp. was 10 degree celcius

Expert's answer

1) As a mathematical equation, Boyle's law is:

P

where:

V - volume;

P - pressure

from this equation the volume will be:

V

V

Answer: 1000 cc

2) p1 = p1

v1= v1

t1=10c =283k

p2=(1-75/100)*p1

v2=(1-40/100)*v1

t2=?

p1*v1/t1=p2*v2/t2

t2=p2*v2*t1/p1*v1

t2=(1-75/100)*p1 * (1-40/100)*v1 * 283k/ p1*v1

from this:

t2=(1-75/100)* (1-40/100)* 283k;

t2 = -0,74*(-0,39) * 283k = 81.67 k

Answer: 81.67 k

P

_{1}V_{1}= P_{2}V_{2}where:

V - volume;

P - pressure

from this equation the volume will be:

V

_{1}= P_{2}V_{2}/P_{1}V

_{1}= 1480 mm * 500cc / 740 mm = 1000 cc;Answer: 1000 cc

2) p1 = p1

v1= v1

t1=10c =283k

p2=(1-75/100)*p1

v2=(1-40/100)*v1

t2=?

p1*v1/t1=p2*v2/t2

t2=p2*v2*t1/p1*v1

t2=(1-75/100)*p1 * (1-40/100)*v1 * 283k/ p1*v1

from this:

t2=(1-75/100)* (1-40/100)* 283k;

t2 = -0,74*(-0,39) * 283k = 81.67 k

Answer: 81.67 k

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