Question

a gas has a volume of 3.80 L. at 0 C. what final temperature, in degrees Celsius, is needed to change the vine of the gas to each of the following, if n and P do not change?

1. 1.40 L (express in 3 Sig fig) 
T1=
2. 1800 L (2 Sig fig)
T2=
3. 12.0 L (3 Sig fig)
T3=
4. 42.0 L (3 Sig fig) 
T4=

Answer 1

We know the

Charle’s law

For at constant pressure and mole

V1/T₁ = V₂/T₂

1. Where, V₁= Initial Volume = 3.80 L, V₂= Final Volume =1.40, T₁= Initial Tempreture =0⁰C = 273.15 K,                               T₂= Final Tempreture = ?.

we can write above equation

T₂ = V₂T₁/V₁

substitute the value in above equation

T₂ = 1.40 273.15 / 3.80 = 100.63 K = -172.52⁰C

thus your T₁ = -172.52⁰C

2. Where, V₁ = Initial Volume = 3.80 L, V₂ = Final Volume =1800L, T₁ = Initial Tempreture =0⁰C = 273.15 K,                               T₂ = Final Tempreture = ?.

we can write above equation

T₂ = V₂T₁/V₁

substitute the value in above equation

T₂ = 1800 273.15 / 3.80 = 129386.84 K = 129113.69⁰C

thus your T₂ = 129113.69⁰C

3. Where, V₁ = Initial Volume = 3.80 L, V₂ = Final Volume =12L, T₁ = Initial Tempreture =0⁰C = 273.15 K,                               T₂ = Final Tempreture = ?.

we can write above equation

T₂ = V₂T₁/V₁

substitute the value in above equation

T₂ = 12 273.15 / 3.80 = 862.57 K = 589.42⁰C

thus your T₃ = 589.92⁰C

2. Where, V₁ = Initial Volume = 3.80 L, V₂ = Final Volume = 42L, T₁ = Initial Tempreture =0⁰C = 273.15 K,                               T₂ = Final Tempreture = ?.

we can write above equation

T₂ = V₂T₁/V₁

substitute the value in above equation

T₂ = 42 273.15 / 3.80 = 3019.02 K = 2745.87⁰C

thus your T₄ = 2745.87⁰C