1. Menghitun kesalahan relative untuk interval waktu 30 detik dan 60 detik

a. Menghitung kesalahan relatif tekanan (h)

Untuk interval waktu 30 detik

h0 = 15 mm = 0,015 m

Δh0 = 1/2 Nst AlatUkur

= ½ x 1 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,015

x 100%

= 3,333 % (3 AP)

(h0 ± Δh0) = (1,50 ± 0,05) X 10-2 m

h1 = 13 mm = 0,013 m

Δh1 = 1/2 Nst AlatUkur

= ½ x 1 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,013

x 100%

= 3,846 % (3 AP)

(h1 ± Δh1) = (1,30 ± 0,05) X 10-2 m

h2 = 12 mm = 0,012 m

Δh2 = 1/2 Nst AlatUkur

= ½ x 2 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,012

x 100%

= 4,167 % (3 AP)

(h2 ± Δh2) = (1,20 ± 0,05) X 10-2 m

h3 = 11 mm = 0,011 m

Δh3 = 1/2 Nst AlatUkur

= ½ x 2 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,011

x 100%

= 4,5454 % (3 AP)

(h3 ± Δh3) = (1,10 ± 0,05) X 10-2 m

h4 = 9 mm = 0,009 m

Δh4 = 1/2 Nst AlatUkur

= ½ x 2 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,009

x 100%

= 5,556 % (2 AP)

(h4 ± Δh4) = (9,0 ± 0,5) X 10-3 m

Untuk interval waktu 60 detik

h0 = 15 mm = 0,015 m

Δh0 = 1/2 Nst AlatUkur

= ½ x 1 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,015

x 100%

= 3,333 % (3 AP)

(h0 ± Δh0) = (1,50 ± 0,05) X 10-2 m

h1 = 13 mm = 0,013 m

Δh1 = 1/2 Nst AlatUkur

= ½ x 1 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,013

x 100%

= 3,846 % (3 AP)

(h1 ± Δh1) = (1,30 ± 0,05) X 10-2 m

h2 = 11 mm = 0,011 m

Δh2 = 1/2 Nst AlatUkur

= ½ x 2 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,011

x 100%

= 4,5454 % (3 AP)

(h2 ± Δh2) = (1,10 ± 0,05) X 10-2 m

h3 = 9 mm = 0,009 m

Δh3 = 1/2 Nst AlatUkur

= ½ x 2 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,009

x 100%

= 5,556 % (2 AP)

(h3 ± Δh3) = (9,0 ± 0,5) X 10-3 m

h4 = 6 mm = 0,006 m

Δh4 = 1/2 Nst AlatUkur

= ½ x 2 mm

= 0,5 mm = 0,0005 m

KR = 0,00050,006

x 100%

= 8,333 % (2 AP)

(h4 ± Δh4) = (6,0 ± 0,5) X 10-3 m

b. Menghitung kesalahan relatif temperature (T)

Untuk interval waktu 30 detik

T0 = 720C

ΔT0 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,572

x 100%

= 0,694 % (3 AP)

(T0 ± ΔT0) = (7,20 ± 5,00) x 10 0C

T1 =620C

ΔT1 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,562

x 100%

= 0,8064 % (3 AP)

(T1 ± ΔT1) = (6,20 ± 5,00)x 10 0C

T2 =560C

ΔT2 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,556

x 100%

= 0,892 % (3 AP)

(T2 ± ΔT2) = (5,60 ± 5,00)x 10 0C

T3 = 520C

ΔT3 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,552

x 100%

= 0,961 % (3 AP)

(T3 ± ΔT3) = (5,20 ± 5,00)x 10 0C

T4 =450C

ΔT4 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,545

x 100%

= 1,111 % (3 AP)

(T4 ± ΔT4) = (4,50 ± 5,00)x 10 0C

Untuk interval waktu 60 detik

T0 = 720C

ΔT0 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,572

x 100%

= 0,694 % (3 AP)

(T0 ± ΔT0) = (7,20 ± 5,00) x 100C

T1 = 640C

ΔT1 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,564

x 100%

= 0,781 % (3 AP)

(T1 ± ΔT1) = (6,40 ± 5,00) x 100C

T2 =560C

ΔT2 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,556

x 100%

= 0,892 % (3 AP)

(T2 ± ΔT2) = (5,60 ± 5,00)x 10 0C

T3 =480C

ΔT3 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,548

x 100%

= 1,0416 % (3 AP)

(T3 ± ΔT3) = (4,80 ± 5,00)x 10 0C

T4 =400C

ΔT4 = 1/2 Nst AlatUkur

= ½ x 10C

=0,5 0C

KR = 0,540

x 100%

= 1,25 % (3 AP)

(T4 ± ΔT4) = (4,00 ± 5,00)x 10 0C

2. Menentukan volume gas pada posisi yang berbeda-beda

Untuk interval waktu 30 detik

h0 = 15 mm = 0,015 m

Δh0 = 1/2 Nst AlatUkur

= ½ x 1 mm = 0,5 mm

=0,0005 m

D =32,5 mm = 0,0325 m

V0 = ¼ π D2 h0

= ¼ x 3,14 x 0,00105625 x 0,015

=0,00001243734375 m3

Δ V0 = 0,00050,015

x 0,00001243734375

= 0,000000414578125 m3

KR = 0,0000004145781250,00001243734375

x 100%

= 3,333% (3 AP)

(V0 ± ΔV0) = (1,24 ±0,41 ) x 10-6 m3

h1 = 13 mm = 0,013 m

Δh1 = 1/2 Nst AlatUkur

= ½ x 1 mm = 0,5 mm

=0,0005 m

D =32,5 mm = 0,0325 m

V1 = ¼ π D2 h0

= ¼ x 3,14 x 0,00105625 x 0,013

= 0,00001077903125m3

Δ V1 = 0,00050,013

x 0,00001077903125

= 0,000000414578125 m3

KR = 0,0000004145781250,00001077903125 5

x 100%

= 3,846 % (3 AP)

(V1 ± ΔV1) = (1,07 ±0,41 ) x 10-6 m3

h2 = 12 mm = 0,012 m

Δh2 = 1/2 Nst AlatUkur

= ½ x 1 mm = 0,5 mm

=0,0005 m

D =32,5 mm = 0,0325 m

V2 = ¼ π D2 h0

= ¼ x 3,14 x 0,00105625 x 0,012

=0,000009949875 m3

Δ V2 = 0,00050,012

x 0,000009949875

= 0,000000414578125 m3

KR = 0,000000414578125

0,000009949875 x 100%

= 4,167 % (3 AP)

(V2 ± ΔV2) = (9,94 ±0,41 ) x 10-6 m3

h3 = 11 mm = 0,011 m

Δh3 = 1/2 Nst AlatUkur

= ½ x 1 mm = 0,5 mm

=0,0005 m

D =32,5 mm = 0,0325 m

V3 = ¼ π D2 h0

= ¼ x 3,14 x 0,00105625 x 0,011

= 0,00000912071875 m3

Δ V3 = 0,00050,011

x 0,00000912071875

= 0,000000414578125 m3

KR = 0,0000004145781250,00000912071875

x 100%

= 4,5454 % (3 AP)

(V3 ± ΔV3) = (9,12 ±0,41 ) x 10-6 m3

h4 = 9 mm = 0,009 m

Δh4 = 1/2 Nst AlatUkur

= ½ x 1 mm = 0,5 mm

=0,0005 m

D =32,5 mm = 0,0325 m

V4 = ¼ π D2 h0

= ¼ x 3,14 x 0,00105625 x 0,009

= 0,00000746240625 m3

Δ V4 = 0,00050,009

x 0,00000746240625

= 0,000000414578125 m3

KR = 0,0000004145781250,00000746240625

x 100%

= 5,556 % (2 AP)

(V4 ± ΔV4) = (7,4 ±0,4 ) x 10-6 m3

Untuk interval waktu 60 detik

h0 = … satuan

Δh0 = 1/2 Nst AlatUkur

= … satuan

D = … satuan m

V0 = ¼ π D2 h0

= … satuan

Δ V0 = ∆ h0

h0 x V0

= … satuan

Δ V0 = ∆ h0

h0

x V0

= …% (…AP)

(V0 ± ΔV0) = (… ± …) satuan

Untuk menghitung nilai volume gas selanjutnya caranya sama seperti yang

di atas

3. Table Hasil Pengamatan

Untuk interval waktu 30 detik

(h0 ± Δh0) satuan (T0 ± ΔT0) satuan (V0 ± ΔV0) satuan

Untuk interval waktu 60 detik

(h0 ± Δh0) satuan (T0 ± ΔT0) satuan (V0 ± ΔV0) satuan

Grafik hubungan antara temperature (T) dan volume (V)

Untuk interval waktu 30 detik

T

V

Interpretasi grafik :

Untuk interval waktu 30 detik

T

V

Interpretasi grafik :

4. Kesimpulan

5. Kemungkinan Kesalahan