Question

In: Chemistry

a gas has a volume of 3.80 L. at 0 C. what final temperature, in degrees...

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=

Solutions

Expert Solution

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


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