371457
In case of an adiabatic process the correct relation in terms of pressure \(p\) and density \(\rho\) of a gas is:-
1 \(p \rho^{\gamma}=\) constant
2 \(p^{\gamma} \rho^{\gamma-1}\) constant
3 \(p \rho^{\gamma-1}=\) constant
4 \(p \rho^{-\gamma}=\) constant
Explanation:
For adiabatic process \(p V^{\gamma}=\) constant \(\begin{aligned}& p\left(\dfrac{M}{\rho}\right)^{\gamma}=\text { constant } \\& p \cdot M^{\gamma} \cdot \rho^{-\gamma}=\text { constant } \\& \text { p. } \rho^{-\gamma}=\text { constant }\end{aligned}\)
PHXI12:THERMODYNAMICS
371458
The temperature of one mole of diatomic gas changes from \(4 T\) to \(T\) in adiabatic process. If \(R\) is universal gas constant. Then work done
1 \(\dfrac{15 R T}{3}\)
2 \(\dfrac{15 R T}{2}\)
3 \(\dfrac{3 R T}{15}\)
4 \(\dfrac{12 R T}{5}\)
Explanation:
\(W=\dfrac{R}{\gamma-1}\left[T_{2}-T_{1}\right]=\dfrac{R}{\dfrac{7}{5}-1}(4 T-T)\) \(=\dfrac{5 R}{2}(3 T)=\dfrac{15 R T}{2}\)
PHXI12:THERMODYNAMICS
371459
A sample of gas at temperature \({T}\) is adiabatically expanded to double its volume. Adiabatic constant for the gas is \({\gamma=3 / 2}\). The work done by the gas in the process is ( \({\mu=1}\) mole)
1 \({R T[2 \sqrt{2}-1]}\)
2 \({R T[\sqrt{2}-2]}\)
3 \({R T[2-\sqrt{2}]}\)
4 \({R T[1-2 \sqrt{2}]}\)
Explanation:
Given: \({\gamma=3 / 2, V_{1}=V, V_{2}=2 V, T_{1}=T}\) For adiabatic process, \({T V^{\gamma-1}=}\) constant \({\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}}\) \({T\left(\dfrac{V}{2 V}\right)^{3 / 2-1}=T_{2} \Rightarrow T_{2}=T\left(\dfrac{1}{2}\right)^{1 / 2}=\dfrac{T}{\sqrt{2}}}\) Work done in adiabatic process, \({W=\dfrac{n R \Delta T}{1-\gamma}}\) \({W=\dfrac{1 \times R\left(\dfrac{T}{\sqrt{2}}-T\right)}{1-\dfrac{3}{2}}=\dfrac{R T(\sqrt{2}-1) \times 2}{\sqrt{2} \times 1}}\) \({\therefore W=R T(2-\sqrt{2})}\) So, correct option is (3).
JEE - 2024
PHXI12:THERMODYNAMICS
371460
During an adiabatic process, the density of a gas is found to be proportional to the cube of temperature. The degree of freedom of gas molecule is
371457
In case of an adiabatic process the correct relation in terms of pressure \(p\) and density \(\rho\) of a gas is:-
1 \(p \rho^{\gamma}=\) constant
2 \(p^{\gamma} \rho^{\gamma-1}\) constant
3 \(p \rho^{\gamma-1}=\) constant
4 \(p \rho^{-\gamma}=\) constant
Explanation:
For adiabatic process \(p V^{\gamma}=\) constant \(\begin{aligned}& p\left(\dfrac{M}{\rho}\right)^{\gamma}=\text { constant } \\& p \cdot M^{\gamma} \cdot \rho^{-\gamma}=\text { constant } \\& \text { p. } \rho^{-\gamma}=\text { constant }\end{aligned}\)
PHXI12:THERMODYNAMICS
371458
The temperature of one mole of diatomic gas changes from \(4 T\) to \(T\) in adiabatic process. If \(R\) is universal gas constant. Then work done
1 \(\dfrac{15 R T}{3}\)
2 \(\dfrac{15 R T}{2}\)
3 \(\dfrac{3 R T}{15}\)
4 \(\dfrac{12 R T}{5}\)
Explanation:
\(W=\dfrac{R}{\gamma-1}\left[T_{2}-T_{1}\right]=\dfrac{R}{\dfrac{7}{5}-1}(4 T-T)\) \(=\dfrac{5 R}{2}(3 T)=\dfrac{15 R T}{2}\)
PHXI12:THERMODYNAMICS
371459
A sample of gas at temperature \({T}\) is adiabatically expanded to double its volume. Adiabatic constant for the gas is \({\gamma=3 / 2}\). The work done by the gas in the process is ( \({\mu=1}\) mole)
1 \({R T[2 \sqrt{2}-1]}\)
2 \({R T[\sqrt{2}-2]}\)
3 \({R T[2-\sqrt{2}]}\)
4 \({R T[1-2 \sqrt{2}]}\)
Explanation:
Given: \({\gamma=3 / 2, V_{1}=V, V_{2}=2 V, T_{1}=T}\) For adiabatic process, \({T V^{\gamma-1}=}\) constant \({\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}}\) \({T\left(\dfrac{V}{2 V}\right)^{3 / 2-1}=T_{2} \Rightarrow T_{2}=T\left(\dfrac{1}{2}\right)^{1 / 2}=\dfrac{T}{\sqrt{2}}}\) Work done in adiabatic process, \({W=\dfrac{n R \Delta T}{1-\gamma}}\) \({W=\dfrac{1 \times R\left(\dfrac{T}{\sqrt{2}}-T\right)}{1-\dfrac{3}{2}}=\dfrac{R T(\sqrt{2}-1) \times 2}{\sqrt{2} \times 1}}\) \({\therefore W=R T(2-\sqrt{2})}\) So, correct option is (3).
JEE - 2024
PHXI12:THERMODYNAMICS
371460
During an adiabatic process, the density of a gas is found to be proportional to the cube of temperature. The degree of freedom of gas molecule is
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PHXI12:THERMODYNAMICS
371457
In case of an adiabatic process the correct relation in terms of pressure \(p\) and density \(\rho\) of a gas is:-
1 \(p \rho^{\gamma}=\) constant
2 \(p^{\gamma} \rho^{\gamma-1}\) constant
3 \(p \rho^{\gamma-1}=\) constant
4 \(p \rho^{-\gamma}=\) constant
Explanation:
For adiabatic process \(p V^{\gamma}=\) constant \(\begin{aligned}& p\left(\dfrac{M}{\rho}\right)^{\gamma}=\text { constant } \\& p \cdot M^{\gamma} \cdot \rho^{-\gamma}=\text { constant } \\& \text { p. } \rho^{-\gamma}=\text { constant }\end{aligned}\)
PHXI12:THERMODYNAMICS
371458
The temperature of one mole of diatomic gas changes from \(4 T\) to \(T\) in adiabatic process. If \(R\) is universal gas constant. Then work done
1 \(\dfrac{15 R T}{3}\)
2 \(\dfrac{15 R T}{2}\)
3 \(\dfrac{3 R T}{15}\)
4 \(\dfrac{12 R T}{5}\)
Explanation:
\(W=\dfrac{R}{\gamma-1}\left[T_{2}-T_{1}\right]=\dfrac{R}{\dfrac{7}{5}-1}(4 T-T)\) \(=\dfrac{5 R}{2}(3 T)=\dfrac{15 R T}{2}\)
PHXI12:THERMODYNAMICS
371459
A sample of gas at temperature \({T}\) is adiabatically expanded to double its volume. Adiabatic constant for the gas is \({\gamma=3 / 2}\). The work done by the gas in the process is ( \({\mu=1}\) mole)
1 \({R T[2 \sqrt{2}-1]}\)
2 \({R T[\sqrt{2}-2]}\)
3 \({R T[2-\sqrt{2}]}\)
4 \({R T[1-2 \sqrt{2}]}\)
Explanation:
Given: \({\gamma=3 / 2, V_{1}=V, V_{2}=2 V, T_{1}=T}\) For adiabatic process, \({T V^{\gamma-1}=}\) constant \({\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}}\) \({T\left(\dfrac{V}{2 V}\right)^{3 / 2-1}=T_{2} \Rightarrow T_{2}=T\left(\dfrac{1}{2}\right)^{1 / 2}=\dfrac{T}{\sqrt{2}}}\) Work done in adiabatic process, \({W=\dfrac{n R \Delta T}{1-\gamma}}\) \({W=\dfrac{1 \times R\left(\dfrac{T}{\sqrt{2}}-T\right)}{1-\dfrac{3}{2}}=\dfrac{R T(\sqrt{2}-1) \times 2}{\sqrt{2} \times 1}}\) \({\therefore W=R T(2-\sqrt{2})}\) So, correct option is (3).
JEE - 2024
PHXI12:THERMODYNAMICS
371460
During an adiabatic process, the density of a gas is found to be proportional to the cube of temperature. The degree of freedom of gas molecule is
371457
In case of an adiabatic process the correct relation in terms of pressure \(p\) and density \(\rho\) of a gas is:-
1 \(p \rho^{\gamma}=\) constant
2 \(p^{\gamma} \rho^{\gamma-1}\) constant
3 \(p \rho^{\gamma-1}=\) constant
4 \(p \rho^{-\gamma}=\) constant
Explanation:
For adiabatic process \(p V^{\gamma}=\) constant \(\begin{aligned}& p\left(\dfrac{M}{\rho}\right)^{\gamma}=\text { constant } \\& p \cdot M^{\gamma} \cdot \rho^{-\gamma}=\text { constant } \\& \text { p. } \rho^{-\gamma}=\text { constant }\end{aligned}\)
PHXI12:THERMODYNAMICS
371458
The temperature of one mole of diatomic gas changes from \(4 T\) to \(T\) in adiabatic process. If \(R\) is universal gas constant. Then work done
1 \(\dfrac{15 R T}{3}\)
2 \(\dfrac{15 R T}{2}\)
3 \(\dfrac{3 R T}{15}\)
4 \(\dfrac{12 R T}{5}\)
Explanation:
\(W=\dfrac{R}{\gamma-1}\left[T_{2}-T_{1}\right]=\dfrac{R}{\dfrac{7}{5}-1}(4 T-T)\) \(=\dfrac{5 R}{2}(3 T)=\dfrac{15 R T}{2}\)
PHXI12:THERMODYNAMICS
371459
A sample of gas at temperature \({T}\) is adiabatically expanded to double its volume. Adiabatic constant for the gas is \({\gamma=3 / 2}\). The work done by the gas in the process is ( \({\mu=1}\) mole)
1 \({R T[2 \sqrt{2}-1]}\)
2 \({R T[\sqrt{2}-2]}\)
3 \({R T[2-\sqrt{2}]}\)
4 \({R T[1-2 \sqrt{2}]}\)
Explanation:
Given: \({\gamma=3 / 2, V_{1}=V, V_{2}=2 V, T_{1}=T}\) For adiabatic process, \({T V^{\gamma-1}=}\) constant \({\therefore T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}}\) \({T\left(\dfrac{V}{2 V}\right)^{3 / 2-1}=T_{2} \Rightarrow T_{2}=T\left(\dfrac{1}{2}\right)^{1 / 2}=\dfrac{T}{\sqrt{2}}}\) Work done in adiabatic process, \({W=\dfrac{n R \Delta T}{1-\gamma}}\) \({W=\dfrac{1 \times R\left(\dfrac{T}{\sqrt{2}}-T\right)}{1-\dfrac{3}{2}}=\dfrac{R T(\sqrt{2}-1) \times 2}{\sqrt{2} \times 1}}\) \({\therefore W=R T(2-\sqrt{2})}\) So, correct option is (3).
JEE - 2024
PHXI12:THERMODYNAMICS
371460
During an adiabatic process, the density of a gas is found to be proportional to the cube of temperature. The degree of freedom of gas molecule is