According to Charles' law, \(\mathrm{\dfrac{V}{T}=K(}\) constant \(\mathrm{)}\) \(\mathrm{\log V-\log T=\log K}\) or \(\mathrm{\log V=\log K+\log T}\) or \(\frac{{{\rm{d(lnV)}}}}{{{\rm{dT}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\;T}}}}\)
KCET - 2019
CHXI06:STATES OF MATTER
314363
The slope of the graph between \(\mathrm{\log \mathrm{V}}\) and \(\mathrm{\log \mathrm{T}}\) (Kelvin scale), for a given mass of a gas is
1 \({\rm{ + 1}}\)
2 \({\rm{ - 1}}\)
3 \(\mathrm{\dfrac{1}{P}}\)
4 \(\mathrm{\dfrac{1}{n}}\)
Explanation:
CHXI06:STATES OF MATTER
314364
A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?
According to Charles' law, \(\mathrm{\dfrac{V}{T}=K(}\) constant \(\mathrm{)}\) \(\mathrm{\log V-\log T=\log K}\) or \(\mathrm{\log V=\log K+\log T}\) or \(\frac{{{\rm{d(lnV)}}}}{{{\rm{dT}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\;T}}}}\)
KCET - 2019
CHXI06:STATES OF MATTER
314363
The slope of the graph between \(\mathrm{\log \mathrm{V}}\) and \(\mathrm{\log \mathrm{T}}\) (Kelvin scale), for a given mass of a gas is
1 \({\rm{ + 1}}\)
2 \({\rm{ - 1}}\)
3 \(\mathrm{\dfrac{1}{P}}\)
4 \(\mathrm{\dfrac{1}{n}}\)
Explanation:
CHXI06:STATES OF MATTER
314364
A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?
According to Charles' law, \(\mathrm{\dfrac{V}{T}=K(}\) constant \(\mathrm{)}\) \(\mathrm{\log V-\log T=\log K}\) or \(\mathrm{\log V=\log K+\log T}\) or \(\frac{{{\rm{d(lnV)}}}}{{{\rm{dT}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\;T}}}}\)
KCET - 2019
CHXI06:STATES OF MATTER
314363
The slope of the graph between \(\mathrm{\log \mathrm{V}}\) and \(\mathrm{\log \mathrm{T}}\) (Kelvin scale), for a given mass of a gas is
1 \({\rm{ + 1}}\)
2 \({\rm{ - 1}}\)
3 \(\mathrm{\dfrac{1}{P}}\)
4 \(\mathrm{\dfrac{1}{n}}\)
Explanation:
CHXI06:STATES OF MATTER
314364
A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?
According to Charles' law, \(\mathrm{\dfrac{V}{T}=K(}\) constant \(\mathrm{)}\) \(\mathrm{\log V-\log T=\log K}\) or \(\mathrm{\log V=\log K+\log T}\) or \(\frac{{{\rm{d(lnV)}}}}{{{\rm{dT}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\;T}}}}\)
KCET - 2019
CHXI06:STATES OF MATTER
314363
The slope of the graph between \(\mathrm{\log \mathrm{V}}\) and \(\mathrm{\log \mathrm{T}}\) (Kelvin scale), for a given mass of a gas is
1 \({\rm{ + 1}}\)
2 \({\rm{ - 1}}\)
3 \(\mathrm{\dfrac{1}{P}}\)
4 \(\mathrm{\dfrac{1}{n}}\)
Explanation:
CHXI06:STATES OF MATTER
314364
A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?