272699
In adiabatic conditions, 2 mole of $\mathrm{CO}_2$ gas at $300 \mathrm{~K}$ is expanded such that its volume becomes 27 times. Then, the work done is $\left(\mathrm{C}_{\mathrm{v}}=6 \mathrm{cal} \mathrm{mol}^{-1}\right.$ and $\left.\gamma=1.33\right)$
272701
A gas expands from the volume of $1 \mathrm{~m}^3$ to a volume of $2 \mathrm{~m}^3$ against an external pressure of $10^5 \mathrm{Nm}^{-2}$. The work done by the gas will be
4 For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
Explanation:
First law of Thermodynamics is $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ (a.) For isochoric process, $\Delta \mathrm{V}=0$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=0$ $\therefore \quad \Delta \mathrm{E}=\mathrm{q}$ (b.) For adiabatic process, $q=0$ $\Delta \mathrm{E}=\mathrm{W}$ (c.) For isothermal process, $\Delta \mathrm{T}=0 \\Delta \mathrm{E}=0$ $\therefore \quad \mathrm{q}=-\mathrm{W}$ (d.) For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
2011
Thermodynamics
272705
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant respectively, then the molar mass of the ideal ga is given by :
1 $\frac{\rho R T}{P}$
2 $\frac{\rho T}{P R}$
3 $\frac{P}{\rho R T}$
4 $\frac{R T}{\rho P}$
Explanation:
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant then ideal gas equation. $P V=n R T$ for $n$ mole of gas $P V=R T$ for 1 mole of gas We know that, $M=V \times \rho \Rightarrow V=\frac{M}{\rho}$ Put the value of $\mathrm{V}$ in gas equation, $P \times \frac{M}{\rho}=\mathrm{RT}$ Then molar mass of gas, $M=\frac{\rho R T}{P}$
Manipal-2019
Thermodynamics
272707
Three moles of an ideal gas are expanded isothermally from a volume of $300 \mathrm{~cm}^3$ to $2.5 \mathrm{~L}$ at $300 \mathrm{~K}$ against a pressure of $1.9 \mathrm{~atm}$. The work done in joules is
272699
In adiabatic conditions, 2 mole of $\mathrm{CO}_2$ gas at $300 \mathrm{~K}$ is expanded such that its volume becomes 27 times. Then, the work done is $\left(\mathrm{C}_{\mathrm{v}}=6 \mathrm{cal} \mathrm{mol}^{-1}\right.$ and $\left.\gamma=1.33\right)$
272701
A gas expands from the volume of $1 \mathrm{~m}^3$ to a volume of $2 \mathrm{~m}^3$ against an external pressure of $10^5 \mathrm{Nm}^{-2}$. The work done by the gas will be
4 For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
Explanation:
First law of Thermodynamics is $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ (a.) For isochoric process, $\Delta \mathrm{V}=0$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=0$ $\therefore \quad \Delta \mathrm{E}=\mathrm{q}$ (b.) For adiabatic process, $q=0$ $\Delta \mathrm{E}=\mathrm{W}$ (c.) For isothermal process, $\Delta \mathrm{T}=0 \\Delta \mathrm{E}=0$ $\therefore \quad \mathrm{q}=-\mathrm{W}$ (d.) For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
2011
Thermodynamics
272705
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant respectively, then the molar mass of the ideal ga is given by :
1 $\frac{\rho R T}{P}$
2 $\frac{\rho T}{P R}$
3 $\frac{P}{\rho R T}$
4 $\frac{R T}{\rho P}$
Explanation:
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant then ideal gas equation. $P V=n R T$ for $n$ mole of gas $P V=R T$ for 1 mole of gas We know that, $M=V \times \rho \Rightarrow V=\frac{M}{\rho}$ Put the value of $\mathrm{V}$ in gas equation, $P \times \frac{M}{\rho}=\mathrm{RT}$ Then molar mass of gas, $M=\frac{\rho R T}{P}$
Manipal-2019
Thermodynamics
272707
Three moles of an ideal gas are expanded isothermally from a volume of $300 \mathrm{~cm}^3$ to $2.5 \mathrm{~L}$ at $300 \mathrm{~K}$ against a pressure of $1.9 \mathrm{~atm}$. The work done in joules is
272699
In adiabatic conditions, 2 mole of $\mathrm{CO}_2$ gas at $300 \mathrm{~K}$ is expanded such that its volume becomes 27 times. Then, the work done is $\left(\mathrm{C}_{\mathrm{v}}=6 \mathrm{cal} \mathrm{mol}^{-1}\right.$ and $\left.\gamma=1.33\right)$
272701
A gas expands from the volume of $1 \mathrm{~m}^3$ to a volume of $2 \mathrm{~m}^3$ against an external pressure of $10^5 \mathrm{Nm}^{-2}$. The work done by the gas will be
4 For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
Explanation:
First law of Thermodynamics is $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ (a.) For isochoric process, $\Delta \mathrm{V}=0$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=0$ $\therefore \quad \Delta \mathrm{E}=\mathrm{q}$ (b.) For adiabatic process, $q=0$ $\Delta \mathrm{E}=\mathrm{W}$ (c.) For isothermal process, $\Delta \mathrm{T}=0 \\Delta \mathrm{E}=0$ $\therefore \quad \mathrm{q}=-\mathrm{W}$ (d.) For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
2011
Thermodynamics
272705
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant respectively, then the molar mass of the ideal ga is given by :
1 $\frac{\rho R T}{P}$
2 $\frac{\rho T}{P R}$
3 $\frac{P}{\rho R T}$
4 $\frac{R T}{\rho P}$
Explanation:
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant then ideal gas equation. $P V=n R T$ for $n$ mole of gas $P V=R T$ for 1 mole of gas We know that, $M=V \times \rho \Rightarrow V=\frac{M}{\rho}$ Put the value of $\mathrm{V}$ in gas equation, $P \times \frac{M}{\rho}=\mathrm{RT}$ Then molar mass of gas, $M=\frac{\rho R T}{P}$
Manipal-2019
Thermodynamics
272707
Three moles of an ideal gas are expanded isothermally from a volume of $300 \mathrm{~cm}^3$ to $2.5 \mathrm{~L}$ at $300 \mathrm{~K}$ against a pressure of $1.9 \mathrm{~atm}$. The work done in joules is
272699
In adiabatic conditions, 2 mole of $\mathrm{CO}_2$ gas at $300 \mathrm{~K}$ is expanded such that its volume becomes 27 times. Then, the work done is $\left(\mathrm{C}_{\mathrm{v}}=6 \mathrm{cal} \mathrm{mol}^{-1}\right.$ and $\left.\gamma=1.33\right)$
272701
A gas expands from the volume of $1 \mathrm{~m}^3$ to a volume of $2 \mathrm{~m}^3$ against an external pressure of $10^5 \mathrm{Nm}^{-2}$. The work done by the gas will be
4 For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
Explanation:
First law of Thermodynamics is $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ (a.) For isochoric process, $\Delta \mathrm{V}=0$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=0$ $\therefore \quad \Delta \mathrm{E}=\mathrm{q}$ (b.) For adiabatic process, $q=0$ $\Delta \mathrm{E}=\mathrm{W}$ (c.) For isothermal process, $\Delta \mathrm{T}=0 \\Delta \mathrm{E}=0$ $\therefore \quad \mathrm{q}=-\mathrm{W}$ (d.) For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
2011
Thermodynamics
272705
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant respectively, then the molar mass of the ideal ga is given by :
1 $\frac{\rho R T}{P}$
2 $\frac{\rho T}{P R}$
3 $\frac{P}{\rho R T}$
4 $\frac{R T}{\rho P}$
Explanation:
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant then ideal gas equation. $P V=n R T$ for $n$ mole of gas $P V=R T$ for 1 mole of gas We know that, $M=V \times \rho \Rightarrow V=\frac{M}{\rho}$ Put the value of $\mathrm{V}$ in gas equation, $P \times \frac{M}{\rho}=\mathrm{RT}$ Then molar mass of gas, $M=\frac{\rho R T}{P}$
Manipal-2019
Thermodynamics
272707
Three moles of an ideal gas are expanded isothermally from a volume of $300 \mathrm{~cm}^3$ to $2.5 \mathrm{~L}$ at $300 \mathrm{~K}$ against a pressure of $1.9 \mathrm{~atm}$. The work done in joules is
272699
In adiabatic conditions, 2 mole of $\mathrm{CO}_2$ gas at $300 \mathrm{~K}$ is expanded such that its volume becomes 27 times. Then, the work done is $\left(\mathrm{C}_{\mathrm{v}}=6 \mathrm{cal} \mathrm{mol}^{-1}\right.$ and $\left.\gamma=1.33\right)$
272701
A gas expands from the volume of $1 \mathrm{~m}^3$ to a volume of $2 \mathrm{~m}^3$ against an external pressure of $10^5 \mathrm{Nm}^{-2}$. The work done by the gas will be
4 For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
Explanation:
First law of Thermodynamics is $\Delta \mathrm{E}=\mathrm{q}+\mathrm{W}$ (a.) For isochoric process, $\Delta \mathrm{V}=0$ $\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=0$ $\therefore \quad \Delta \mathrm{E}=\mathrm{q}$ (b.) For adiabatic process, $q=0$ $\Delta \mathrm{E}=\mathrm{W}$ (c.) For isothermal process, $\Delta \mathrm{T}=0 \\Delta \mathrm{E}=0$ $\therefore \quad \mathrm{q}=-\mathrm{W}$ (d.) For cyclic process, state functions like $\Delta \mathrm{E}=0$ $\mathrm{q}=-\mathrm{W}$
2011
Thermodynamics
272705
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant respectively, then the molar mass of the ideal ga is given by :
1 $\frac{\rho R T}{P}$
2 $\frac{\rho T}{P R}$
3 $\frac{P}{\rho R T}$
4 $\frac{R T}{\rho P}$
Explanation:
If $P, T, \rho$ and $R$ represents pressure, temperature, density and universal gas constant then ideal gas equation. $P V=n R T$ for $n$ mole of gas $P V=R T$ for 1 mole of gas We know that, $M=V \times \rho \Rightarrow V=\frac{M}{\rho}$ Put the value of $\mathrm{V}$ in gas equation, $P \times \frac{M}{\rho}=\mathrm{RT}$ Then molar mass of gas, $M=\frac{\rho R T}{P}$
Manipal-2019
Thermodynamics
272707
Three moles of an ideal gas are expanded isothermally from a volume of $300 \mathrm{~cm}^3$ to $2.5 \mathrm{~L}$ at $300 \mathrm{~K}$ against a pressure of $1.9 \mathrm{~atm}$. The work done in joules is
